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Multiple Angles Class 10 Optional Math | Trigonometry
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1. Multiple Angles: Class 10 Optional Mathematics Trigonometry Overview and Key Formulas

Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.2 Multiple Angles

Multiple angles in trigonometry are angles such as \(2A\), \(3A\) and \(4A\), which are whole-number multiples of an angle \(A\). This page gives the multiple angles formula (double angle and triple angle), the way to get each trigonometry multiple angle formula from the compound angle formulas, and step by step solutions to every question of Exercise 7.2 (Class 10 Optional Mathematics, SEE, Nepal), with the Nepali question in brackets.

This is part of the unit prescribed by the Curriculum Development Centre (CDC), Nepal. Multiple choice solutions also have Nepali in brackets, and the other solutions are in English because the steps and symbols are the same in both languages.

  • Multiple angles (Exercise 7.2): put \(B=A\) to get the double angle formulas, then expand \(A+2A\) for the triple angle formulas.

Key formulas to remember

Double angle: \(\sin2A=2\sin A\cos A\), \(\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A\), \(\tan2A=\dfrac{2\tan A}{1-\tan^2A}\).
Triple angle: \(\sin3A=3\sin A-4\sin^3A\), \(\cos3A=4\cos^3A-3\cos A\), \(\tan3A=\dfrac{3\tan A-\tan^3A}{1-3\tan^2A}\). Also \(1+\cos2A=2\cos^2A\) and \(1-\cos2A=2\sin^2A\).

For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about trigonometric identities for extra understanding.

2. Exercise 7.2 Multiple Angles: Class 10 Optional Mathematics Trigonometry

Exercise 7.2 Multiple Angles: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)

Q.1 Tick (✓) the correct option for the given questions:

(दिइएका प्रश्नहरूको उपयुक्त विकल्पमा ठिक चिह्न (✓) लगाउनुहोस् : )

A. Which of the following trigonometric expressions is equal to \(\cos2A\)? (तलका मध्ये कुन \(\cos2A\) सँग बराबर हुन्छ ?)

  • a) \(\cos^2A+\sin^2A\)
  • b) \(\cos^2A-\sin^2A\)
  • c) \(1-2\cos^2A\)
  • d) \(2\sin^2A-1\)

Correct Answer: b) \(\cos^2A-\sin^2A\)

Solution:

  • Put \(B=A\) in \(\cos(A+B)=\cos A\cos B-\sin A\sin B\).
\[\begin{aligned}\cos2A &= \cos(A+A) \\ &= \cos A\cos A-\sin A\sin A \\ &= \cos^2A-\sin^2A\end{aligned}\]

(\(\cos(A+B)\) मा \(B=A\) राख्दा \(\cos2A=\cos^2A-\sin^2A\) हुन्छ ।)

B. Which of the following trigonometric expressions is equal to \(\sin3\theta\)? (तलका मध्ये कुन \(\sin3\theta\) सँग बराबर हुन्छ ?)

  • a) \(4\sin\theta-3\sin^3\theta\)
  • b) \(4\sin^3\theta-3\sin\theta\)
  • c) \(3\sin\theta-4\sin^3\theta\)
  • d) \(3\sin\theta-4\sin^4\theta\)

Correct Answer: c) \(3\sin\theta-4\sin^3\theta\)

Solution:

  • Expand \(\sin(\theta+2\theta)\).
\[\begin{aligned}\sin3\theta &= \sin\theta\cos2\theta+\cos\theta\sin2\theta \\ &= \sin\theta(1-2\sin^2\theta)+2\sin\theta\cos^2\theta \\ &= 3\sin\theta-4\sin^3\theta\end{aligned}\]

(\(\sin3\theta=3\sin\theta-4\sin^3\theta\) हुन्छ ।)

C. If \(A=30^\circ\), what is the value of \(\sin2A\)? (यदि \(A=30^\circ\) भए \(\sin2A\) को मान कति हुन्छ ?)

  • a) \(0\)
  • b) \(1\)
  • c) \(\frac12\)
  • d) \(\frac{\sqrt3}{2}\)

Correct Answer: d) \(\frac{\sqrt3}{2}\)

Solution:

  • Here, \(\sin2A=\sin(2\times30^\circ)=\sin60^\circ=\frac{\sqrt3}{2}\).

(\(\sin2A=\sin60^\circ=\frac{\sqrt3}{2}\) हुन्छ ।)

D. What should be the value of \(A\) when \(\cos3A=-1\)? (\(\cos3A=-1\) हुँदा \(A\) को मान कति हुनुपर्छ ?)

  • a) \(0^\circ\)
  • b) \(30^\circ\)
  • c) \(15^\circ\)
  • d) \(60^\circ\)

Correct Answer: d) \(60^\circ\)

Solution:

  • Since \(\cos180^\circ=-1\), we get \(3A=180^\circ\).
\[A=\frac{180^\circ}{3}=60^\circ\]

(\(3A=180^\circ\) भएकाले \(A=60^\circ\) हुन्छ ।)

E. Which of the following trigonometric expressions is equal to \(\tan2A\)? (तलका मध्ये कुन \(\tan2A\) सँग बराबर हुन्छ ?)

  • a) \(\frac{2\tan A}{1-\tan^2A}\)
  • b) \(\frac{2\tan A}{1+\tan^2A}\)
  • c) \(\frac{1+\tan^2A}{1-\tan^2A}\)
  • d) \(\frac{1-\tan^2A}{1+\tan^2A}\)

Correct Answer: a) \(\frac{2\tan A}{1-\tan^2A}\)

Solution:

  • Put \(B=A\) in \(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\).
\[\begin{aligned}\tan2A &= \frac{\tan A+\tan A}{1-\tan A\tan A} \\ &= \frac{2\tan A}{1-\tan^2A}\end{aligned}\]

(\(\tan2A=\frac{2\tan A}{1-\tan^2A}\) हुन्छ ।)

Exercise 7.2 Multiple Angles: Written Questions (लेख्ने प्रश्न)

Q.2
a. If \(\sin A=\frac45\), find the values of \(\sin2A\), \(\cos2A\) and \(\tan2A\). (यदि \(\sin A=\frac45\) भए \(\sin2A\), \(\cos2A\) र \(\tan2A\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\sin A=\frac45\) (taking \(A\) acute).

\[\begin{aligned}\cos A &= \sqrt{1-\sin^2A} \\ &= \sqrt{1-\frac{16}{25}} \\ &= \frac35\end{aligned}\]\[\begin{aligned}\sin2A &= 2\sin A\cos A \\ &= 2\cdot\frac45\cdot\frac35 \\ &= \frac{24}{25}\end{aligned}\]\[\begin{aligned}\cos2A &= 1-2\sin^2A \\ &= 1-2\cdot\frac{16}{25} \\ &= -\frac{7}{25}\end{aligned}\]\[\begin{aligned}\tan2A &= \frac{\sin2A}{\cos2A} \\ &= \frac{24/25}{-7/25} \\ &= -\frac{24}{7}\end{aligned}\]
\(\sin2A=\dfrac{24}{25},\ \cos2A=-\dfrac{7}{25},\ \tan2A=-\dfrac{24}{7}\)
b. If \(\cos A=\frac{12}{13}\), find the values of \(\sin2A\), \(\cos2A\) and \(\tan2A\). (यदि \(\cos A=\frac{12}{13}\) भए \(\sin2A\), \(\cos2A\) र \(\tan2A\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\cos A=\frac{12}{13}\) (taking \(A\) acute).

\[\begin{aligned}\sin A &= \sqrt{1-\cos^2A} \\ &= \sqrt{1-\frac{144}{169}} \\ &= \frac{5}{13}\end{aligned}\]\[\begin{aligned}\sin2A &= 2\sin A\cos A \\ &= 2\cdot\frac{5}{13}\cdot\frac{12}{13} \\ &= \frac{120}{169}\end{aligned}\]\[\begin{aligned}\cos2A &= 2\cos^2A-1 \\ &= 2\cdot\frac{144}{169}-1 \\ &= \frac{119}{169}\end{aligned}\]\[\begin{aligned}\tan2A &= \frac{120/169}{119/169} \\ &= \frac{120}{119}\end{aligned}\]
\(\sin2A=\dfrac{120}{169},\ \cos2A=\dfrac{119}{169},\ \tan2A=\dfrac{120}{119}\)
c. If \(\cos\theta=\frac45\), find the values of \(\sin3\theta\), \(\cos3\theta\) and \(\tan3\theta\). (यदि \(\cos\theta=\frac45\) भए \(\sin3\theta\), \(\cos3\theta\) र \(\tan3\theta\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\cos\theta=\frac45\) (taking \(\theta\) acute), so \(\sin\theta=\sqrt{1-\frac{16}{25}}=\frac35\).

\[\begin{aligned}\sin3\theta &= 3\sin\theta-4\sin^3\theta \\ &= \frac95-4\cdot\frac{27}{125} \\ &= \frac{225-108}{125} \\ &= \frac{117}{125}\end{aligned}\]\[\begin{aligned}\cos3\theta &= 4\cos^3\theta-3\cos\theta \\ &= 4\cdot\frac{64}{125}-\frac{12}{5} \\ &= \frac{256-300}{125} \\ &= -\frac{44}{125}\end{aligned}\]\[\begin{aligned}\tan3\theta &= \frac{117/125}{-44/125} \\ &= -\frac{117}{44}\end{aligned}\]
\(\sin3\theta=\dfrac{117}{125},\ \cos3\theta=-\dfrac{44}{125},\ \tan3\theta=-\dfrac{117}{44}\)
d. If \(\tan y=\frac12\), find the value of \(\tan3y\). (यदि \(\tan y=\frac12\) भए \(\tan3y\) को मान पत्ता लगाउनुहोस् ।)

Use \(\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}\).

\[\begin{aligned}\tan3y &= \frac{\frac32-\frac18}{1-\frac34} \\ &= \frac{11/8}{1/4} \\ &= \frac{11}{2}\end{aligned}\]
\(\tan3y=\dfrac{11}{2}\)
Q.3
a. If \(\cos2A=\frac{7}{25}\), find the value of \(\sin A\). (यदि \(\cos2A=\frac{7}{25}\) भए \(\sin A\) को मान पत्ता लगाउनुहोस् ।)

Use \(\cos2A=1-2\sin^2A\) (taking \(A\) acute).

\[\begin{aligned}& \frac{7}{25}=1-2\sin^2A \\ &\implies 2\sin^2A=\frac{18}{25} \\ &\implies \sin^2A=\frac{9}{25} \\ &\implies \sin A=\frac35\end{aligned}\]
\(\sin A=\dfrac35\)
b. If \(\cos2A=\frac{13}{36}\), find the value of \(\cos A\). (यदि \(\cos2A=\frac{13}{36}\) भए \(\cos A\) को मान पत्ता लगाउनुहोस् ।)

Use \(\cos2A=2\cos^2A-1\) (taking \(A\) acute).

\[\begin{aligned}& \frac{13}{36}=2\cos^2A-1 \\ &\implies 2\cos^2A=\frac{49}{36} \\ &\implies \cos^2A=\frac{49}{72} \\ &\implies \cos A=\frac{7}{6\sqrt2}\end{aligned}\]
\(\cos A=\dfrac{7}{6\sqrt2}\)
c. If \(\tan2A=\frac{120}{119}\), find the value of \(\tan A\). (यदि \(\tan2A=\frac{120}{119}\) भए \(\tan A\) को मान पत्ता लगाउनुहोस् ।)

Let \(t=\tan A\). Use \(\tan2A=\dfrac{2t}{1-t^2}\).

\[\begin{aligned}& \frac{120}{119}=\frac{2t}{1-t^2} \\ &\implies 120-120t^2=238t \\ &\implies 60t^2+119t-60=0\end{aligned}\]

Factorize:

\[\begin{aligned}& 60t^2+144t-25t-60=0 \\ &\implies 12t(5t+12)-5(5t+12)=0 \\ &\implies (12t-5)(5t+12)=0\end{aligned}\]
\(\tan A=\dfrac{5}{12}\) or \(\tan A=-\dfrac{12}{5}\)

Q.4 Prove that:

  • a. \(\frac{1+\cos2A}{\sin2A}=\cot A\)
  • b. \(\frac{1-\cos2A}{\sin2A}=\tan A\)
  • c. \(\frac{1-\cos2A}{1+\cos2A}=\tan^2A\)
  • d. \(\frac{1-\tan\theta}{1+\tan\theta}=\frac{1-\sin2\theta}{\cos2\theta}\)
  • e. \(\frac{\cos2\alpha}{1+\sin2\alpha}=\frac{1-\tan\alpha}{1+\tan\alpha}\)
  • f. \(\frac{1+\sin2\theta-\cos2\theta}{1+\sin2\theta+\cos2\theta}=\tan\theta\)
  • g. \(\frac{\sin\theta+\sin2\theta}{1+\cos\theta+\cos2\theta}=\tan\theta\)
  • h. \(\tan A+\cot A=2\csc2A\)
  • i. \(\cot A-\tan A=2\cot2A\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{1+\cos2A}{\sin2A}=\cot A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1+(2\cos^2A-1)}{2\sin A\cos A} \\ &= \frac{2\cos^2A}{2\sin A\cos A} \\ &= \frac{\cos A}{\sin A} \\ &= \cot A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{1-\cos2A}{\sin2A}=\tan A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1-(1-2\sin^2A)}{2\sin A\cos A} \\ &= \frac{2\sin^2A}{2\sin A\cos A} \\ &= \frac{\sin A}{\cos A} \\ &= \tan A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\frac{1-\cos2A}{1+\cos2A}=\tan^2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{2\sin^2A}{2\cos^2A} \\ &= \tan^2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\frac{1-\tan\theta}{1+\tan\theta}=\frac{1-\sin2\theta}{\cos2\theta}\)
\[\begin{aligned}\text{R.H.S.} &= \frac{1-\sin2\theta}{\cos2\theta} \\ &= \frac{\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} \\ &= \frac{(\cos\theta-\sin\theta)^2}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)} \\ &= \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} \\ &= \frac{1-\tan\theta}{1+\tan\theta} \\ &= \text{L.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\frac{\cos2\alpha}{1+\sin2\alpha}=\frac{1-\tan\alpha}{1+\tan\alpha}\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha+2\sin\alpha\cos\alpha} \\ &= \frac{(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)}{(\cos\alpha+\sin\alpha)^2} \\ &= \frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha} \\ &= \frac{1-\tan\alpha}{1+\tan\alpha} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\frac{1+\sin2\theta-\cos2\theta}{1+\sin2\theta+\cos2\theta}=\tan\theta\)
\[\begin{aligned}\text{L.H.S.} &= \frac{(1-\cos2\theta)+\sin2\theta}{(1+\cos2\theta)+\sin2\theta} \\ &= \frac{2\sin^2\theta+2\sin\theta\cos\theta}{2\cos^2\theta+2\sin\theta\cos\theta} \\ &= \frac{2\sin\theta(\sin\theta+\cos\theta)}{2\cos\theta(\cos\theta+\sin\theta)} \\ &= \tan\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
g. \(\frac{\sin\theta+\sin2\theta}{1+\cos\theta+\cos2\theta}=\tan\theta\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin\theta+\sin2\theta}{(1+\cos2\theta)+\cos\theta} \\ &= \frac{\sin\theta+2\sin\theta\cos\theta}{2\cos^2\theta+\cos\theta} \\ &= \frac{\sin\theta(1+2\cos\theta)}{\cos\theta(2\cos\theta+1)} \\ &= \tan\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
h. \(\tan A+\cot A=2\csc2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \\ &= \frac{\sin^2A+\cos^2A}{\sin A\cos A} \\ &= \frac{1}{\sin A\cos A} \\ &= \frac{2}{2\sin A\cos A} \\ &= \frac{2}{\sin2A} \\ &= 2\csc2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
i. \(\cot A-\tan A=2\cot2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos A}{\sin A}-\frac{\sin A}{\cos A} \\ &= \frac{\cos^2A-\sin^2A}{\sin A\cos A} \\ &= \frac{2\cos2A}{2\sin A\cos A} \\ &= \frac{2\cos2A}{\sin2A} \\ &= 2\cot2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.5 Prove that:

  • a. \(\tan(45^\circ+A)=\frac{\cos2A}{1-\sin2A}\)
  • b. \(\frac{1-\sin2\theta}{\cos2\theta}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\)
  • c. \(2\sin^2(45^\circ-A)=1-\sin2A\)
  • d. \(\tan(45^\circ+A)=\sec2A+\tan2A\)
  • e. \(2\cos^2(45^\circ-A)=1+\sin2A\)
  • f. \(\cos^2(45^\circ-\theta)-\sin^2(45^\circ-\theta)=\sin2\theta\)
  • g. \(\frac{\sin5A}{\sin A}-\frac{\cos5A}{\cos A}=4\cos2A\)
  • h. \(\frac{1-\tan^2(45^\circ-A)}{1+\tan^2(45^\circ-A)}=\sin2A\)

(प्रमाणित गर्नुहोस् : )

a. \(\tan(45^\circ+A)=\frac{\cos2A}{1-\sin2A}\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1+\tan A}{1-\tan A} \\ &= \frac{\cos A+\sin A}{\cos A-\sin A} \\ &= \frac{(\cos A+\sin A)(\cos A-\sin A)}{(\cos A-\sin A)^2} \\ &= \frac{\cos^2A-\sin^2A}{\cos^2A+\sin^2A-2\sin A\cos A} \\ &= \frac{\cos2A}{1-\sin2A} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{1-\sin2\theta}{\cos2\theta}=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} \\ &= \frac{(\cos\theta-\sin\theta)^2}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)} \\ &= \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(2\sin^2(45^\circ-A)=1-\sin2A\)

Using \(2\sin^2X=1-\cos2X\):

\[\begin{aligned}\text{L.H.S.} &= 1-\cos2(45^\circ-A) \\ &= 1-\cos(90^\circ-2A) \\ &= 1-\sin2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\tan(45^\circ+A)=\sec2A+\tan2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1+\tan A}{1-\tan A} \\ &= \frac{\cos A+\sin A}{\cos A-\sin A} \\ &= \frac{(\cos A+\sin A)^2}{\cos^2A-\sin^2A} \\ &= \frac{1+\sin2A}{\cos2A} \\ &= \frac{1}{\cos2A}+\frac{\sin2A}{\cos2A} \\ &= \sec2A+\tan2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(2\cos^2(45^\circ-A)=1+\sin2A\)

Using \(2\cos^2X=1+\cos2X\):

\[\begin{aligned}\text{L.H.S.} &= 1+\cos2(45^\circ-A) \\ &= 1+\cos(90^\circ-2A) \\ &= 1+\sin2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\cos^2(45^\circ-\theta)-\sin^2(45^\circ-\theta)=\sin2\theta\)

Using \(\cos^2X-\sin^2X=\cos2X\):

\[\begin{aligned}\text{L.H.S.} &= \cos2(45^\circ-\theta) \\ &= \cos(90^\circ-2\theta) \\ &= \sin2\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
g. \(\frac{\sin5A}{\sin A}-\frac{\cos5A}{\cos A}=4\cos2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin5A\cos A-\cos5A\sin A}{\sin A\cos A} \\ &= \frac{\sin4A}{\sin A\cos A} \\ &= \frac{2\sin4A}{\sin2A} \\ &= \frac{2(2\sin2A\cos2A)}{\sin2A} \\ &= 4\cos2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
h. \(\frac{1-\tan^2(45^\circ-A)}{1+\tan^2(45^\circ-A)}=\sin2A\)

Using \(\dfrac{1-\tan^2X}{1+\tan^2X}=\cos2X\):

\[\begin{aligned}\text{L.H.S.} &= \cos2(45^\circ-A) \\ &= \cos(90^\circ-2A) \\ &= \sin2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
Q.6
a. If \(\cos\theta=\frac12\left(a+\frac1a\right)\), prove that: \(\cos2\theta=\frac12\left(a^2+\frac{1}{a^2}\right)\) (यदि \(\cos\theta=\frac12\left(a+\frac1a\right)\) भए प्रमाणित गर्नुहोस् : \(\cos2\theta=\frac12\left(a^2+\frac{1}{a^2}\right)\))

Given \(\cos\theta=\frac12\left(a+\frac1a\right)\).

\[\begin{aligned}\cos2\theta &= 2\cos^2\theta-1 \\ &= 2\cdot\frac14\left(a+\frac1a\right)^2-1 \\ &= \frac12\left(a^2+2+\frac{1}{a^2}\right)-1 \\ &= \frac12\left(a^2+\frac{1}{a^2}\right)\end{aligned}\]
Hence proved.
b. If \(\sin A=\frac12\left(a+\frac1a\right)\), prove that: \(\cos2A=-\frac12\left(a^2+\frac{1}{a^2}\right)\) (यदि \(\sin A=\frac12\left(a+\frac1a\right)\) भए प्रमाणित गर्नुहोस् : \(\cos2A=-\frac12\left(a^2+\frac{1}{a^2}\right)\))

Given \(\sin A=\frac12\left(a+\frac1a\right)\).

\[\begin{aligned}\cos2A &= 1-2\sin^2A \\ &= 1-2\cdot\frac14\left(a+\frac1a\right)^2 \\ &= 1-\frac12\left(a^2+2+\frac{1}{a^2}\right) \\ &= -\frac12\left(a^2+\frac{1}{a^2}\right)\end{aligned}\]
Hence proved.
c. If \(\sin\beta=\frac12\left(a+\frac1a\right)\), prove that: \(\sin3\beta=-\frac12\left(a^3+\frac{1}{a^3}\right)\) (यदि \(\sin\beta=\frac12\left(a+\frac1a\right)\) भए प्रमाणित गर्नुहोस् : \(\sin3\beta=-\frac12\left(a^3+\frac{1}{a^3}\right)\))

Given \(\sin\beta=\frac12\left(a+\frac1a\right)\), so \(\sin^3\beta=\frac18\left(a^3+3a+\frac3a+\frac{1}{a^3}\right)\).

\[\begin{aligned}\sin3\beta &= 3\sin\beta-4\sin^3\beta \\ &= \frac32\left(a+\frac1a\right)-\frac12\left(a^3+3a+\frac3a+\frac{1}{a^3}\right) \\ &= \frac32\left(a+\frac1a\right)-\frac12\left(a^3+\frac{1}{a^3}\right)-\frac32\left(a+\frac1a\right) \\ &= -\frac12\left(a^3+\frac{1}{a^3}\right)\end{aligned}\]
Hence proved.
d. If \(\cos\beta=\frac12\left(a+\frac1a\right)\), prove that: \(\cos3\beta=\frac12\left(a^3+\frac{1}{a^3}\right)\) (यदि \(\cos\beta=\frac12\left(a+\frac1a\right)\) भए प्रमाणित गर्नुहोस् : \(\cos3\beta=\frac12\left(a^3+\frac{1}{a^3}\right)\))

Given \(\cos\beta=\frac12\left(a+\frac1a\right)\), so \(\cos^3\beta=\frac18\left(a^3+3a+\frac3a+\frac{1}{a^3}\right)\).

\[\begin{aligned}\cos3\beta &= 4\cos^3\beta-3\cos\beta \\ &= \frac12\left(a^3+3a+\frac3a+\frac{1}{a^3}\right)-\frac32\left(a+\frac1a\right) \\ &= \frac12\left(a^3+\frac{1}{a^3}\right)+\frac32\left(a+\frac1a\right)-\frac32\left(a+\frac1a\right) \\ &= \frac12\left(a^3+\frac{1}{a^3}\right)\end{aligned}\]
Hence proved.

Q.7 Prove that:

  • a. \((\sin2A+\sin2B)^2+(\cos2A-\cos2B)^2=4\sin^2(A+B)\)
  • b. \((\sin2A-\sin2B)^2+(\cos2A+\cos2B)^2=4\cos^2(A+B)\)
  • c. \((1+\sin2A+\cos2A)^2=4\cos^2A(1+\sin2A)\)

(प्रमाणित गर्नुहोस् : )

a. \((\sin2A+\sin2B)^2+(\cos2A-\cos2B)^2=4\sin^2(A+B)\)
\[\begin{aligned}\text{L.H.S.} &= \sin^22A+2\sin2A\sin2B+\sin^22B+\cos^22A-2\cos2A\cos2B+\cos^22B \\ &= (\sin^22A+\cos^22A)+(\sin^22B+\cos^22B)-2(\cos2A\cos2B-\sin2A\sin2B) \\ &= 2-2\cos(2A+2B) \\ &= 2[1-\cos2(A+B)] \\ &= 2\cdot2\sin^2(A+B) \\ &= 4\sin^2(A+B) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \((\sin2A-\sin2B)^2+(\cos2A+\cos2B)^2=4\cos^2(A+B)\)
\[\begin{aligned}\text{L.H.S.} &= (\sin^22A+\cos^22A)+(\sin^22B+\cos^22B)+2(\cos2A\cos2B-\sin2A\sin2B) \\ &= 2+2\cos(2A+2B) \\ &= 2[1+\cos2(A+B)] \\ &= 2\cdot2\cos^2(A+B) \\ &= 4\cos^2(A+B) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \((1+\sin2A+\cos2A)^2=4\cos^2A(1+\sin2A)\)
\[\begin{aligned}\text{L.H.S.} &= [(1+\cos2A)+\sin2A]^2 \\ &= (2\cos^2A+2\sin A\cos A)^2 \\ &= [2\cos A(\cos A+\sin A)]^2 \\ &= 4\cos^2A(\cos^2A+2\sin A\cos A+\sin^2A) \\ &= 4\cos^2A(1+\sin2A) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.8 Prove that:

  • a. \(\sin^2A-\cos^2A\cos2B=\sin^2B-\cos^2B\cos2A\)
  • b. \(\cos^2\alpha+\sin^2\alpha\cos2\beta=\cos^2\beta+\sin^2\beta\cos2\alpha\)

(प्रमाणित गर्नुहोस् : )

a. \(\sin^2A-\cos^2A\cos2B=\sin^2B-\cos^2B\cos2A\)

Let \(s=\sin^2A\) and \(t=\sin^2B\). Using \(\cos2B=1-2t\) and \(\cos^2A=1-s\):

\[\begin{aligned}\text{L.H.S.} &= s-(1-s)(1-2t) \\ &= s-(1-2t-s+2st) \\ &= 2s+2t-2st-1\end{aligned}\]

Using \(\cos2A=1-2s\) and \(\cos^2B=1-t\):

\[\begin{aligned}\text{R.H.S.} &= t-(1-t)(1-2s) \\ &= t-(1-2s-t+2st) \\ &= 2s+2t-2st-1\end{aligned}\]

Both sides are equal to \(2\sin^2A+2\sin^2B-2\sin^2A\sin^2B-1\).

Therefore, L.H.S. = R.H.S., proved.
b. \(\cos^2\alpha+\sin^2\alpha\cos2\beta=\cos^2\beta+\sin^2\beta\cos2\alpha\)

Let \(c=\cos^2\alpha\) and \(d=\cos^2\beta\). Using \(\sin^2\alpha=1-c\) and \(\cos2\beta=2d-1\):

\[\begin{aligned}\text{L.H.S.} &= c+(1-c)(2d-1) \\ &= c+2d-1-2cd+c \\ &= 2c+2d-2cd-1\end{aligned}\]

Using \(\sin^2\beta=1-d\) and \(\cos2\alpha=2c-1\):

\[\begin{aligned}\text{R.H.S.} &= d+(1-d)(2c-1) \\ &= d+2c-1-2cd+d \\ &= 2c+2d-2cd-1\end{aligned}\]

Both sides are equal.

Therefore, L.H.S. = R.H.S., proved.

Q.9 Prove that:

  • a. \((2\cos\theta+1)(2\cos\theta-1)=2\cos2\theta+1\)
  • b. \((2\cos\theta+1)(2\cos\theta-1)(2\cos2\theta-1)=2\cos4\theta+1\)

(प्रमाणित गर्नुहोस् : )

a. \((2\cos\theta+1)(2\cos\theta-1)=2\cos2\theta+1\)
\[\begin{aligned}\text{L.H.S.} &= 4\cos^2\theta-1 \\ &= 2(2\cos^2\theta)-1 \\ &= 2(1+\cos2\theta)-1 \\ &= 2\cos2\theta+1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \((2\cos\theta+1)(2\cos\theta-1)(2\cos2\theta-1)=2\cos4\theta+1\)

Using part (a), \((2\cos\theta+1)(2\cos\theta-1)=2\cos2\theta+1\).

\[\begin{aligned}\text{L.H.S.} &= (2\cos2\theta+1)(2\cos2\theta-1) \\ &= 4\cos^22\theta-1 \\ &= 2(1+\cos4\theta)-1 \\ &= 2\cos4\theta+1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.10 Prove that:

  • a. \(\frac{\cos^3A+\sin^3A}{\cos A+\sin A}=1-\frac12\sin2A\)
  • b. \(\cos^6\theta-\sin^6\theta=\cos2\theta\left(1-\frac14\sin^22\theta\right)\)
  • c. \(4(\sin^6\theta+\cos^6\theta)=4-3\sin^22\theta\)
  • d. \(\cos^6\theta-\sin^6\theta=\frac14(\cos^32\theta+3\cos2\theta)\)
  • e. \(\cos^6\phi+\sin^6\phi=\frac18(5+3\cos4\phi)\)
  • f. \(\sin^8\theta+\cos^8\theta=\frac18(8-8\sin^22\theta+\sin^42\theta)\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\cos^3A+\sin^3A}{\cos A+\sin A}=1-\frac12\sin2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{(\cos A+\sin A)(\cos^2A-\sin A\cos A+\sin^2A)}{\cos A+\sin A} \\ &= 1-\sin A\cos A \\ &= 1-\frac12\sin2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\cos^6\theta-\sin^6\theta=\cos2\theta\left(1-\frac14\sin^22\theta\right)\)
\[\begin{aligned}\text{L.H.S.} &= (\cos^2\theta-\sin^2\theta)(\cos^4\theta+\cos^2\theta\sin^2\theta+\sin^4\theta) \\ &= \cos2\theta[(\cos^2\theta+\sin^2\theta)^2-\sin^2\theta\cos^2\theta] \\ &= \cos2\theta\left(1-\frac14(2\sin\theta\cos\theta)^2\right) \\ &= \cos2\theta\left(1-\frac14\sin^22\theta\right) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(4(\sin^6\theta+\cos^6\theta)=4-3\sin^22\theta\)

Using \(a^3+b^3=(a+b)^3-3ab(a+b)\) with \(a=\sin^2\theta\) and \(b=\cos^2\theta\):

\[\begin{aligned}\text{L.H.S.} &= 4[1-3\sin^2\theta\cos^2\theta] \\ &= 4-3(2\sin\theta\cos\theta)^2 \\ &= 4-3\sin^22\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\cos^6\theta-\sin^6\theta=\frac14(\cos^32\theta+3\cos2\theta)\)

From part (b) and \(\sin^22\theta=1-\cos^22\theta\):

\[\begin{aligned}\text{L.H.S.} &= \cos2\theta\left(1-\frac14\sin^22\theta\right) \\ &= \cos2\theta\left(1-\frac14+\frac14\cos^22\theta\right) \\ &= \frac14\cos2\theta(3+\cos^22\theta) \\ &= \frac14(\cos^32\theta+3\cos2\theta) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\cos^6\phi+\sin^6\phi=\frac18(5+3\cos4\phi)\)

From part (c), \(\cos^6\phi+\sin^6\phi=1-\frac34\sin^22\phi\) and \(\sin^22\phi=\dfrac{1-\cos4\phi}{2}\).

\[\begin{aligned}\text{L.H.S.} &= 1-\frac34\cdot\frac{1-\cos4\phi}{2} \\ &= \frac{8-3+3\cos4\phi}{8} \\ &= \frac18(5+3\cos4\phi) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\sin^8\theta+\cos^8\theta=\frac18(8-8\sin^22\theta+\sin^42\theta)\)

Let \(x=\sin^22\theta\). Then \(\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta=1-\frac{x}{2}\) and \(\sin^4\theta\cos^4\theta=\frac{x^2}{16}\).

\[\begin{aligned}\text{L.H.S.} &= (\sin^4\theta+\cos^4\theta)^2-2\sin^4\theta\cos^4\theta \\ &= \left(1-\frac x2\right)^2-\frac{x^2}{8} \\ &= 1-x+\frac{x^2}{4}-\frac{x^2}{8} \\ &= \frac{8-8x+x^2}{8} \\ &= \frac18(8-8\sin^22\theta+\sin^42\theta) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.11 Prove that:

  • a. \(\csc10^\circ-\sqrt3\sec10^\circ=4\)
  • b. \(\sqrt3\csc20^\circ-\sec20^\circ=4\)
  • c. \(\sec40^\circ+\sqrt3\csc40^\circ=4\)

(प्रमाणित गर्नुहोस् : )

a. \(\csc10^\circ-\sqrt3\sec10^\circ=4\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ} \\ &= \frac{2\left(\frac12\cos10^\circ-\frac{\sqrt3}{2}\sin10^\circ\right)}{\frac12\sin20^\circ} \\ &= \frac{2(\sin30^\circ\cos10^\circ-\cos30^\circ\sin10^\circ)}{\frac12\sin20^\circ} \\ &= \frac{2\sin20^\circ}{\frac12\sin20^\circ} \\ &= 4 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\sqrt3\csc20^\circ-\sec20^\circ=4\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sqrt3\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ} \\ &= \frac{2(\sin60^\circ\cos20^\circ-\cos60^\circ\sin20^\circ)}{\frac12\sin40^\circ} \\ &= \frac{2\sin40^\circ}{\frac12\sin40^\circ} \\ &= 4 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\sec40^\circ+\sqrt3\csc40^\circ=4\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin40^\circ+\sqrt3\cos40^\circ}{\sin40^\circ\cos40^\circ} \\ &= \frac{2(\cos60^\circ\sin40^\circ+\sin60^\circ\cos40^\circ)}{\frac12\sin80^\circ} \\ &= \frac{2\sin100^\circ}{\frac12\sin80^\circ} \\ &= \frac{2\sin80^\circ}{\frac12\sin80^\circ} \\ &= 4 \\ &= \text{R.H.S.}\end{aligned}\]

Here \(\sin100^\circ=\sin(180^\circ-80^\circ)=\sin80^\circ\).

Therefore, L.H.S. = R.H.S., proved.

Q.12 Prove that:

  • a. \(\csc2A+\cot4A=\cot A-\csc4A\)
  • b. \(\cot8A+\csc4A=\cot2A-\csc8A\)
  • c. \(\frac{\sec4A-1}{\sec2A-1}=\tan4A\cot A\)
  • d. \(\frac{\sec8A-1}{\sec4A-1}=\tan8A\cot2A\)

(प्रमाणित गर्नुहोस् : )

a. \(\csc2A+\cot4A=\cot A-\csc4A\)

Using \(\csc X+\cot X=\dfrac{1+\cos X}{\sin X}=\cot\dfrac X2\):

\[\csc4A+\cot4A=\cot2A \implies \csc2A+\cot4A+\csc4A=\csc2A+\cot2A=\cot A\]

Hence \(\csc2A+\cot4A=\cot A-\csc4A\).

Hence proved.
b. \(\cot8A+\csc4A=\cot2A-\csc8A\)

Using \(\csc X+\cot X=\cot\dfrac X2\):

\[\cot8A+\csc8A=\cot4A \implies \cot8A+\csc8A+\csc4A=\cot4A+\csc4A=\cot2A\]

Hence \(\cot8A+\csc4A=\cot2A-\csc8A\).

Hence proved.
c. \(\frac{\sec4A-1}{\sec2A-1}=\tan4A\cot A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1-\cos4A}{1-\cos2A}\cdot\frac{\cos2A}{\cos4A} \\ &= \frac{2\sin^22A}{2\sin^2A}\cdot\frac{\cos2A}{\cos4A} \\ &= \frac{(2\sin A\cos A)(\sin2A\cos2A)}{\sin^2A\cos4A} \\ &= \frac{(2\sin A\cos A)\left(\frac12\sin4A\right)}{\sin^2A\cos4A} \\ &= \frac{\cos A}{\sin A}\cdot\frac{\sin4A}{\cos4A} \\ &= \cot A\tan4A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\frac{\sec8A-1}{\sec4A-1}=\tan8A\cot2A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1-\cos8A}{1-\cos4A}\cdot\frac{\cos4A}{\cos8A} \\ &= \frac{2\sin^24A}{2\sin^22A}\cdot\frac{\cos4A}{\cos8A} \\ &= \frac{(2\sin2A\cos2A)\left(\frac12\sin8A\right)}{\sin^22A\cos8A} \\ &= \frac{\cos2A}{\sin2A}\cdot\frac{\sin8A}{\cos8A} \\ &= \cot2A\tan8A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.13 Prove that:

  • a. \(\frac{1}{\tan3\theta+\tan\theta}-\frac{1}{\cot3\theta+\cot\theta}=\cot4\theta\)
  • b. \(\frac{\cot\theta}{\cot\theta-\cot3\theta}-\frac{\tan\theta}{\tan3\theta-\tan\theta}=1\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{1}{\tan3\theta+\tan\theta}-\frac{1}{\cot3\theta+\cot\theta}=\cot4\theta\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos3\theta\cos\theta}{\sin3\theta\cos\theta+\cos3\theta\sin\theta}-\frac{\sin3\theta\sin\theta}{\cos3\theta\sin\theta+\sin3\theta\cos\theta} \\ &= \frac{\cos3\theta\cos\theta-\sin3\theta\sin\theta}{\sin4\theta} \\ &= \frac{\cos4\theta}{\sin4\theta} \\ &= \cot4\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{\cot\theta}{\cot\theta-\cot3\theta}-\frac{\tan\theta}{\tan3\theta-\tan\theta}=1\)

First, \(\dfrac{\cot\theta}{\cot\theta-\cot3\theta}=\dfrac{\frac{1}{\tan\theta}}{\frac{\tan3\theta-\tan\theta}{\tan\theta\tan3\theta}}=\dfrac{\tan3\theta}{\tan3\theta-\tan\theta}\).

\[\begin{aligned}\text{L.H.S.} &= \frac{\tan3\theta}{\tan3\theta-\tan\theta}-\frac{\tan\theta}{\tan3\theta-\tan\theta} \\ &= \frac{\tan3\theta-\tan\theta}{\tan3\theta-\tan\theta} \\ &= 1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.14 Prove that:

  • a. \(4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)\)
  • b. \(4(\cos^320^\circ+\sin^350^\circ)=3(\cos20^\circ+\sin50^\circ)\)
  • c. \(\cos^3A\cos3A+\sin^3A\sin3A=\cos^32A\)

(प्रमाणित गर्नुहोस् : )

a. \(4(\cos^310^\circ+\sin^320^\circ)=3(\cos10^\circ+\sin20^\circ)\)

Using \(4\cos^3A=\cos3A+3\cos A\) and \(4\sin^3A=3\sin A-\sin3A\):

\[\begin{aligned}\text{L.H.S.} &= \cos30^\circ+3\cos10^\circ+3\sin20^\circ-\sin60^\circ \\ &= \frac{\sqrt3}{2}+3(\cos10^\circ+\sin20^\circ)-\frac{\sqrt3}{2} \\ &= 3(\cos10^\circ+\sin20^\circ) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(4(\cos^320^\circ+\sin^350^\circ)=3(\cos20^\circ+\sin50^\circ)\)

Using \(4\cos^3A=\cos3A+3\cos A\) and \(4\sin^3A=3\sin A-\sin3A\):

\[\begin{aligned}\text{L.H.S.} &= \cos60^\circ+3\cos20^\circ+3\sin50^\circ-\sin150^\circ \\ &= \frac12+3(\cos20^\circ+\sin50^\circ)-\frac12 \\ &= 3(\cos20^\circ+\sin50^\circ) \\ &= \text{R.H.S.}\end{aligned}\]

Here \(\sin150^\circ=\sin(180^\circ-30^\circ)=\frac12\).

Therefore, L.H.S. = R.H.S., proved.
c. \(\cos^3A\cos3A+\sin^3A\sin3A=\cos^32A\)
\[\begin{aligned}\text{L.H.S.} &= \cos^3A(4\cos^3A-3\cos A)+\sin^3A(3\sin A-4\sin^3A) \\ &= 4(\cos^6A-\sin^6A)-3(\cos^4A-\sin^4A) \\ &= 4\cos2A(1-\sin^2A\cos^2A)-3\cos2A \\ &= \cos2A(1-4\sin^2A\cos^2A) \\ &= \cos2A(1-\sin^22A) \\ &= \cos^32A \\ &= \text{R.H.S.}\end{aligned}\]

Here \(\cos^4A-\sin^4A=\cos2A\) and \(\cos^6A-\sin^6A=\cos2A(1-\sin^2A\cos^2A)\).

Therefore, L.H.S. = R.H.S., proved.

Q.15 Prove that:

  • a. \(\tan\theta+\tan(60^\circ+\theta)-\tan(60^\circ-\theta)=3\tan3\theta\)
  • b. \(\cot\theta+\cot(60^\circ+\theta)-\cot(60^\circ-\theta)=3\cot3\theta\)
  • c. \(\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta=\cot\theta\)

(प्रमाणित गर्नुहोस् : )

a. \(\tan\theta+\tan(60^\circ+\theta)-\tan(60^\circ-\theta)=3\tan3\theta\)

Let \(t=\tan\theta\). Then

\[\begin{aligned}\tan(60^\circ+\theta)-\tan(60^\circ-\theta) &= \frac{\sqrt3+t}{1-\sqrt3t}-\frac{\sqrt3-t}{1+\sqrt3t} \\ &= \frac{(\sqrt3+t)(1+\sqrt3t)-(\sqrt3-t)(1-\sqrt3t)}{1-3t^2} \\ &= \frac{8t}{1-3t^2}\end{aligned}\]\[\begin{aligned}\text{L.H.S.} &= t+\frac{8t}{1-3t^2} \\ &= \frac{t(1-3t^2)+8t}{1-3t^2} \\ &= 3\cdot\frac{3t-t^3}{1-3t^2} \\ &= 3\tan3\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\cot\theta+\cot(60^\circ+\theta)-\cot(60^\circ-\theta)=3\cot3\theta\)

Let \(t=\tan\theta\). Then

\[\begin{aligned}\cot(60^\circ+\theta)-\cot(60^\circ-\theta) &= \frac{1-\sqrt3t}{\sqrt3+t}-\frac{1+\sqrt3t}{\sqrt3-t} \\ &= \frac{(1-\sqrt3t)(\sqrt3-t)-(1+\sqrt3t)(\sqrt3+t)}{3-t^2} \\ &= \frac{-8t}{3-t^2}\end{aligned}\]\[\begin{aligned}\text{L.H.S.} &= \frac1t-\frac{8t}{3-t^2} \\ &= \frac{3-t^2-8t^2}{t(3-t^2)} \\ &= 3\cdot\frac{1-3t^2}{3t-t^3} \\ &= 3\cot3\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta=\cot\theta\)

Using \(\cot X-\tan X=2\cot2X\), that is \(\tan X+2\cot2X=\cot X\):

\[\begin{aligned}8\cot8\theta+4\tan4\theta &= 4(2\cot8\theta+\tan4\theta) \\ &= 4\cot4\theta\end{aligned}\]\[\begin{aligned}4\cot4\theta+2\tan2\theta &= 2(2\cot4\theta+\tan2\theta) \\ &= 2\cot2\theta\end{aligned}\]\[2\cot2\theta+\tan\theta=\cot\theta\]

Hence the sum \(\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta=\cot\theta\).

Hence proved.
Q.16
a. Express \(\cos4\theta\) in terms of \(\sin\theta\). (\(\cos4\theta\) लाई \(\sin\theta\) का रूपमा व्यक्त गर्नुहोस् ।)

Here,

\[\begin{aligned}\cos4\theta &= 1-2\sin^22\theta \\ &= 1-2(2\sin\theta\cos\theta)^2 \\ &= 1-8\sin^2\theta(1-\sin^2\theta) \\ &= 1-8\sin^2\theta+8\sin^4\theta\end{aligned}\]
\(\cos4\theta=8\sin^4\theta-8\sin^2\theta+1\)
b. Express \(\cos5\theta\) in terms of \(\cos\theta\). (\(\cos5\theta\) लाई \(\cos\theta\) का रूपमा व्यक्त गर्नुहोस् ।)

Let \(c=\cos\theta\). Here \(\cos5\theta=\cos3\theta\cos2\theta-\sin3\theta\sin2\theta\).

\[\begin{aligned}\cos3\theta\cos2\theta &= (4c^3-3c)(2c^2-1) \\ &= 8c^5-10c^3+3c\end{aligned}\]\[\begin{aligned}\sin3\theta\sin2\theta &= (3\sin\theta-4\sin^3\theta)(2\sin\theta\cos\theta) \\ &= 2c\,\sin^2\theta(3-4\sin^2\theta) \\ &= 2c(1-c^2)(4c^2-1) \\ &= 10c^3-8c^5-2c\end{aligned}\]

Subtract:

\[\begin{aligned}\cos5\theta &= (8c^5-10c^3+3c)-(10c^3-8c^5-2c) \\ &= 16c^5-20c^3+5c\end{aligned}\]
\(\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta\)

Q.17 Prove that:

  • a. \(\frac{\cos A}{\cos A-\sin A}-\frac{\cos A}{\cos A+\sin A}=\tan2A\)
  • b. \(\tan2\theta+\sec2\theta=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\)
  • c. \(\frac{\cot2\theta+\tan2\theta}{\cot2\theta-\tan2\theta}=\sec4\theta\)
  • d. \(\frac{\cos2A}{1+\sin2A}=\tan(45^\circ-A)=\frac{1-\tan A}{1+\tan A}\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\cos A}{\cos A-\sin A}-\frac{\cos A}{\cos A+\sin A}=\tan2A\)
\[\begin{aligned}\text{L.H.S.} &= \cos A\cdot\frac{(\cos A+\sin A)-(\cos A-\sin A)}{\cos^2A-\sin^2A} \\ &= \frac{2\sin A\cos A}{\cos2A} \\ &= \frac{\sin2A}{\cos2A} \\ &= \tan2A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\tan2\theta+\sec2\theta=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\)
\[\begin{aligned}\text{L.H.S.} &= \frac{1+\sin2\theta}{\cos2\theta} \\ &= \frac{(\cos\theta+\sin\theta)^2}{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)} \\ &= \frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\frac{\cot2\theta+\tan2\theta}{\cot2\theta-\tan2\theta}=\sec4\theta\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos^22\theta+\sin^22\theta}{\cos^22\theta-\sin^22\theta} \\ &= \frac{1}{\cos4\theta} \\ &= \sec4\theta \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\frac{\cos2A}{1+\sin2A}=\tan(45^\circ-A)=\frac{1-\tan A}{1+\tan A}\)

First part:

\[\begin{aligned}\frac{\cos2A}{1+\sin2A} &= \frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A+\sin A)^2} \\ &= \frac{\cos A-\sin A}{\cos A+\sin A} \\ &= \frac{1-\tan A}{1+\tan A}\end{aligned}\]

Second part:

\[\begin{aligned}\tan(45^\circ-A) &= \frac{\tan45^\circ-\tan A}{1+\tan45^\circ\tan A} \\ &= \frac{1-\tan A}{1+\tan A}\end{aligned}\]

So all three expressions are equal.

Hence proved.

Q.18 If \(2\tan\alpha=3\tan\beta\), prove that:

  • a. \(\tan(\alpha+\beta)=\dfrac{5\sin2\beta}{5\cos2\beta-1}\)
  • b. \(\tan(\alpha-\beta)=\dfrac{\sin2\beta}{5-\cos2\beta}\)

(यदि \(2\tan\alpha=3\tan\beta\) भए प्रमाणित गर्नुहोस् : )

a. \(\tan(\alpha+\beta)=\dfrac{5\sin2\beta}{5\cos2\beta-1}\)

Let \(t=\tan\beta\), so \(\tan\alpha=\frac32t\).

\[\begin{aligned}\tan(\alpha+\beta) &= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ &= \frac{\frac52t}{1-\frac32t^2} \\ &= \frac{5t}{2-3t^2}\end{aligned}\]

Now use \(t=\dfrac{\sin2\beta}{1+\cos2\beta}\) and \(t^2=\dfrac{1-\cos2\beta}{1+\cos2\beta}\):

\[\begin{aligned}\frac{5t}{2-3t^2} &= \frac{5\sin2\beta}{2(1+\cos2\beta)-3(1-\cos2\beta)} \\ &= \frac{5\sin2\beta}{5\cos2\beta-1}\end{aligned}\]
Hence proved.
b. \(\tan(\alpha-\beta)=\dfrac{\sin2\beta}{5-\cos2\beta}\)

Let \(t=\tan\beta\), so \(\tan\alpha=\frac32t\).

\[\begin{aligned}\tan(\alpha-\beta) &= \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} \\ &= \frac{\frac12t}{1+\frac32t^2} \\ &= \frac{t}{2+3t^2}\end{aligned}\]

Using \(t=\dfrac{\sin2\beta}{1+\cos2\beta}\) and \(t^2=\dfrac{1-\cos2\beta}{1+\cos2\beta}\):

\[\begin{aligned}\frac{t}{2+3t^2} &= \frac{\sin2\beta}{2(1+\cos2\beta)+3(1-\cos2\beta)} \\ &= \frac{\sin2\beta}{5-\cos2\beta}\end{aligned}\]
Hence proved.

Q.19 If \(\tan A=\frac ba\), prove that: \(a\cos2A+b\sin2A=a\).

(यदि \(\tan A=\frac ba\) भए प्रमाणित गर्नुहोस् : \(a\cos2A+b\sin2A=a\) ।)

Here, \(\tan A=\frac ba\).

\[\begin{aligned}\cos2A &= \frac{1-\tan^2A}{1+\tan^2A} \\ &= \frac{a^2-b^2}{a^2+b^2},\qquad \sin2A \\ &= \frac{2\tan A}{1+\tan^2A} \\ &= \frac{2ab}{a^2+b^2}\end{aligned}\]\[\begin{aligned}\text{L.H.S.} &= a\cdot\frac{a^2-b^2}{a^2+b^2}+b\cdot\frac{2ab}{a^2+b^2} \\ &= \frac{a^3-ab^2+2ab^2}{a^2+b^2} \\ &= \frac{a(a^2+b^2)}{a^2+b^2} \\ &= a \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.20 Prove that: \(\dfrac{\sin A-\sqrt{1-\sin2A}}{\cos A-\sqrt{1-\sin2A}}=\cot A\).

(प्रमाणित गर्नुहोस् : \(\dfrac{\sin A-\sqrt{1-\sin2A}}{\cos A-\sqrt{1-\sin2A}}=\cot A\) ।)

Simplifying the surd

\[1-\sin2A=\sin^2A+\cos^2A-2\sin A\cos A=(\cos A-\sin A)^2 \implies \sqrt{1-\sin2A}=|\cos A-\sin A|\]

Case 1: \(0<A<45^\circ\), so \(\sqrt{1-\sin2A}=\cos A-\sin A\)

\[\begin{aligned}\frac{\sin A-(\cos A-\sin A)}{\cos A-(\cos A-\sin A)} &= \frac{2\sin A-\cos A}{\sin A} \\ &= 2-\cot A\end{aligned}\]

Case 2: \(45^\circ<A<90^\circ\), so \(\sqrt{1-\sin2A}=\sin A-\cos A\)

\[\begin{aligned}\frac{\sin A-(\sin A-\cos A)}{\cos A-(\sin A-\cos A)} &= \frac{\cos A}{2\cos A-\sin A} \\ &= \frac{1}{2-\tan A}\end{aligned}\]

Note on the printed question

Neither case gives \(\cot A\). For \(A=30^\circ\) the left side is \(\dfrac{0.5-0.366}{0.866-0.366}\approx0.268=2-\sqrt3\), while \(\cot30^\circ=\sqrt3\approx1.732\). So the identity cannot be proved as printed; there appears to be a misprint in the question.

The expression simplifies to \(2-\cot A\) for \(0<A<45^\circ\), not \(\cot A\).

Q.21 Prove that: \(\sin^4A=\frac38-\frac12\cos2A+\frac18\cos4A\)

(प्रमाणित गर्नुहोस् : \(\sin^4A=\frac38-\frac12\cos2A+\frac18\cos4A\))

\[\begin{aligned}\text{L.H.S.} &= \left(\frac{1-\cos2A}{2}\right)^2 \\ &= \frac{1-2\cos2A+\cos^22A}{4} \\ &= \frac{1-2\cos2A+\frac{1+\cos4A}{2}}{4} \\ &= \frac{3-4\cos2A+\cos4A}{8} \\ &= \frac38-\frac12\cos2A+\frac18\cos4A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.22 Prove that: \(\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}+\sin^4\frac{5\pi}{8}+\sin^4\frac{7\pi}{8}=\frac32\)

(प्रमाणित गर्नुहोस् : \(\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}+\sin^4\frac{5\pi}{8}+\sin^4\frac{7\pi}{8}=\frac32\))

Since \(\sin\frac{7\pi}{8}=\sin\frac{\pi}{8}\), \(\sin\frac{5\pi}{8}=\sin\frac{3\pi}{8}\) and \(\sin\frac{3\pi}{8}=\cos\frac{\pi}{8}\):

\[\begin{aligned}\text{L.H.S.} &= 2\left[\sin^4\frac{\pi}{8}+\cos^4\frac{\pi}{8}\right] \\ &= 2\left[1-2\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}\right] \\ &= 2\left[1-\frac12\sin^2\frac{\pi}{4}\right] \\ &= 2\left[1-\frac14\right] \\ &= \frac32 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
Answer
1. A. b   B. c   C. d   D. d   E. a
2. a. \(\frac{24}{25},-\frac{7}{25},-\frac{24}{7}\)   b. \(\frac{120}{169},\frac{119}{169},\frac{120}{119}\)   c. \(\frac{117}{125},-\frac{44}{125},-\frac{117}{44}\)   d. \(\frac{11}{2}\)
3. a. \(\frac35\)   b. \(\frac{7}{6\sqrt2}\)   c. \(\frac{5}{12}\) or \(-\frac{12}{5}\)   16. a. \(8\sin^4\theta-8\sin^2\theta+1\)   b. \(16\cos^5\theta-20\cos^3\theta+5\cos\theta\)
4–15, 17–19, 21, 22. Proved   20. See the note (printed identity does not hold)

3. Frequently Asked Questions on Multiple Angles Trigonometry (Class 10 Optional Mathematics)

What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.2 Multiple Angles? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.2 Multiple Angles मा के के समावेश छ ?)

This page solves all 73 questions of Exercise 7.2 Multiple Angles, in textbook order, with the Nepali question in brackets and an Answer box at the end. Multiple angles formula for sin2A, cos2A, tan2A and sin3A, cos3A, tan3A with all Exercise 7.2 questions solved for Class 10 Optional Math SEE Nepal.

What are multiple angles in trigonometry? (त्रिकोणमितिमा बहुगुणित कोण भनेको के हो ?)

A multiple angle is an angle that is a whole-number multiple of another angle, for example \(2A\) and \(3A\). Exercise 7.2 uses the double angle and triple angle formulas to find values from a given ratio and to prove long identities.

What is the multiple angles formula? (बहुगुणित कोणका सूत्रहरू के के हुन् ?)
  • \(\sin2A=2\sin A\cos A\)
  • \(\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A\)
  • \(\tan2A=\dfrac{2\tan A}{1-\tan^2A}\)
  • \(\sin3A=3\sin A-4\sin^3A\), \(\cos3A=4\cos^3A-3\cos A\), \(\tan3A=\dfrac{3\tan A-\tan^3A}{1-3\tan^2A}\)
What are the double angle formulas? (दोब्बर कोणका सूत्रहरू के के हुन् ?)
  • \(\sin2A=2\sin A\cos A\)
  • \(\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A\)
  • \(\tan2A=\dfrac{2\tan A}{1-\tan^2A}\)

They come from the compound angle formulas by putting \(B=A\) (Question 1).

What are the triple angle formulas? (तेब्बर कोणका सूत्रहरू के के हुन् ?)
  • \(\sin3A=3\sin A-4\sin^3A\)
  • \(\cos3A=4\cos^3A-3\cos A\)
  • \(\tan3A=\dfrac{3\tan A-\tan^3A}{1-3\tan^2A}\)
Is multiple angles in trigonometry the same as a multiple angle camera or 3D camera? (के त्रिकोणमितिको बहुगुणित कोण क्यामेराको बहुकोण जस्तै हो ?)

No. A multiple angle camera or 3D camera shows one scene from several viewpoints. In trigonometry a multiple angle is an angle such as \(2A\) or \(3A\), and its formulas express \(\sin2A\), \(\cos3A\) and so on in terms of \(A\).

How do I find \(\sin A\) when \(\cos2A\) is given? (\(\cos2A\) दिइएको छ भने \(\sin A\) कसरी निकाल्ने ?)

Use \(\cos2A=1-2\sin^2A\) and solve for \(\sin A\). For \(\cos2A=\frac{7}{25}\), \(\sin^2A=\frac{9}{25}\), so \(\sin A=\frac35\) (Question 3a).

How do I prove identities that contain \(1\pm\cos2A\)? (\(1\pm\cos2A\) भएका सर्वसमिका कसरी प्रमाणित गर्ने ?)

Use \(1+\cos2A=2\cos^2A\) and \(1-\cos2A=2\sin^2A\), together with \(\sin2A=2\sin A\cos A\). Example: \(\dfrac{1+\cos2A}{\sin2A}=\cot A\) (Question 4a).

Is every printed identity in Exercise 7.2 true? (के अभ्यास ७.२ मा छापिएका सबै सर्वसमिका सही छन् ?)

Not always. Question 20 as printed does not give \(\cot A\): for \(0

Are these Exercise 7.2 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)

Yes. Exercise 7.2 covers finding values from a given ratio, long identities and expressing one angle in terms of another, which are common SEE-style question types in Optional Mathematics.

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