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

Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.1 Compound Angles

Compound angles in trigonometry are angles written as the sum or difference of two angles, such as \(A+B\) or \(A-B\). This page gives the trigonometry compound angle formula, a short compound angle formula proof, compound angles examples and step by step solutions to every compound angles trigonometry question of Exercise 7.1 (Class 10 Optional Mathematics, SEE, Nepal). Every question is written as in the textbook, with its parts listed under it and 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.

  • Compound angles: formulas for the sine, cosine and tangent of \(A+B\) and \(A-B\).
  • Exact values: write \(15^\circ=45^\circ-30^\circ\), \(75^\circ=45^\circ+30^\circ\), \(105^\circ=60^\circ+45^\circ\) and use standard values.
  • Identities: use \(A+B=45^\circ\) (then \(\tan(A+B)=1\)), divide by \(\cos\), and expand \(A+B+C\) step by step.

Key formulas to remember

\(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\) and \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\).
\(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\). Also \(\sin15^\circ=\cos75^\circ=\dfrac{\sqrt3-1}{2\sqrt2}\) and \(\sin75^\circ=\cos15^\circ=\dfrac{\sqrt3+1}{2\sqrt2}\).
Cotangent: \(\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}\) and \(\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}\).
Proof idea for \(\cos(A-B)\): on the unit circle take \(P(\cos A,\sin A)\) and \(Q(\cos B,\sin B)\). By the distance formula \(PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2=2-2(\cos A\cos B+\sin A\sin B)\), and by the cosine rule \(PQ^2=2-2\cos(A-B)\). Equating gives \(\cos(A-B)=\cos A\cos B+\sin A\sin B\); the other formulas follow by putting \(-B\) for \(B\) and using \(\cos(90^\circ-x)=\sin x\).

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

2. Exercise 7.1 Compound Angles: Class 10 Optional Mathematics Trigonometry

Exercise 7.1 Compound 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 one is equal to \(\sin(A+B)\)? (तलका मध्ये कुन \(\sin(A+B)\) सँग बराबर हुन्छ ?)

  • a) \(\sin A\cos B-\cos A\sin B\)
  • b) \(\sin A\cos B+\cos A\sin B\)
  • c) \(\cos A\cos B-\sin A\sin B\)
  • d) \(\sin A\cos A+\cos B\sin B\)

Correct Answer: b) \(\sin A\cos B+\cos A\sin B\)

Solution:

  • By the compound angle formula for the sine of a sum:
\[\sin(A+B)=\sin A\cos B+\cos A\sin B\]

(योगफलको साइनको सूत्र अनुसार \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) हुन्छ ।)

B. Which is equal to \(\cos(A-B)\)? (तलका मध्ये कुन \(\cos(A-B)\) सँग बराबर हुन्छ ?)

  • a) \(\sin A\cos B-\cos A\sin B\)
  • b) \(\sin A\cos B+\cos A\sin B\)
  • c) \(\cos A\cos B+\sin A\sin B\)
  • d) \(\sin A\cos A+\cos B\sin B\)

Correct Answer: c) \(\cos A\cos B+\sin A\sin B\)

Solution:

  • By the compound angle formula for the cosine of a difference:
\[\cos(A-B)=\cos A\cos B+\sin A\sin B\]

(अन्तरको कोसाइनको सूत्र अनुसार \(\cos(A-B)=\cos A\cos B+\sin A\sin B\) हुन्छ ।)

C. If \(A=30^\circ\) and \(B=60^\circ\), what is the value of \(\sin(A+B)\)? (यदि \(A=30^\circ\) र \(B=60^\circ\) भए \(\sin(A+B)\) को मान कति हुन्छ ?)

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

Correct Answer: b) \(1\)

Solution:

  • Here, \(A+B=30^\circ+60^\circ=90^\circ\).
\[\sin(A+B)=\sin90^\circ=1\]

(\(A+B=90^\circ\) भएकाले \(\sin90^\circ=1\) हुन्छ ।)

D. If \(\sin(A+B)=\frac12\) and \(B=30^\circ\), what is the value of \(A\)? (यदि \(\sin(A+B)=\frac12\) र \(B=30^\circ\) भए \(A\) को मान कति हुन्छ ?)

  • a) \(0^\circ\)
  • b) \(30^\circ\)
  • c) \(45^\circ\)
  • d) \(60^\circ\)

Correct Answer: a) \(0^\circ\)

Solution:

  • Since \(\sin30^\circ=\frac12\), we get \(A+B=30^\circ\).
\[A+30^\circ=30^\circ \implies A=0^\circ\]

(\(\sin(A+B)=\sin30^\circ\) भएकाले \(A+B=30^\circ\) र \(A=0^\circ\) हुन्छ ।)

E. Which of the following relations is correct? (तलका सम्बन्धहरू मध्ये कुन सही छ ?)

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

Correct Answer: d) \(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)

Solution:

  • The tangent of a sum is
\[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\]

(\(\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\) हुन्छ ।)

Exercise 7.1 Compound Angles: Written Questions (लेख्ने प्रश्न)

Q.2 Prove that: (Without using calculator or table)

  • a. \(\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}\)
  • b. \(\sin75^\circ=\frac{\sqrt3+1}{2\sqrt2}\)
  • c. \(\tan75^\circ=2+\sqrt3\)
  • d. \(\cos105^\circ=\frac{1-\sqrt3}{2\sqrt2}\)
  • e. \(\tan105^\circ=-(2+\sqrt3)\)
  • f. \(\cot15^\circ=2+\sqrt3\)

(क्याल्कुलेटर वा तालिका प्रयोग नगरी सिद्ध गर्नुहोस् : )

a. \(\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}\)

Here, \(15^\circ=45^\circ-30^\circ\) and \(\cos(A-B)=\cos A\cos B+\sin A\sin B\).

\[\begin{aligned}\text{L.H.S.} &= \cos(45^\circ-30^\circ) \\ &= \cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ \\ &= \frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12 \\ &= \frac{\sqrt3+1}{2\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\sin75^\circ=\frac{\sqrt3+1}{2\sqrt2}\)

Here, \(75^\circ=45^\circ+30^\circ\) and \(\sin(A+B)=\sin A\cos B+\cos A\sin B\).

\[\begin{aligned}\text{L.H.S.} &= \sin(45^\circ+30^\circ) \\ &= \sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ \\ &= \frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12 \\ &= \frac{\sqrt3+1}{2\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\tan75^\circ=2+\sqrt3\)

Here, \(75^\circ=45^\circ+30^\circ\).

\[\begin{aligned}\text{L.H.S.} &= \frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ} \\ &= \frac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}} \\ &= \frac{\sqrt3+1}{\sqrt3-1} \\ &= \frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)} \\ &= \frac{4+2\sqrt3}{2} \\ &= 2+\sqrt3 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\cos105^\circ=\frac{1-\sqrt3}{2\sqrt2}\)

Here, \(105^\circ=60^\circ+45^\circ\) and \(\cos(A+B)=\cos A\cos B-\sin A\sin B\).

\[\begin{aligned}\text{L.H.S.} &= \cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ \\ &= \frac12\cdot\frac{1}{\sqrt2}-\frac{\sqrt3}{2}\cdot\frac{1}{\sqrt2} \\ &= \frac{1-\sqrt3}{2\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\tan105^\circ=-(2+\sqrt3)\)

Here, \(105^\circ=60^\circ+45^\circ\).

\[\begin{aligned}\text{L.H.S.} &= \frac{\tan60^\circ+\tan45^\circ}{1-\tan60^\circ\tan45^\circ} \\ &= \frac{\sqrt3+1}{1-\sqrt3} \\ &= \frac{(\sqrt3+1)(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)} \\ &= \frac{4+2\sqrt3}{-2} \\ &= -(2+\sqrt3) \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\cot15^\circ=2+\sqrt3\)

Here, \(\cot15^\circ=\dfrac{1}{\tan15^\circ}\) and \(15^\circ=45^\circ-30^\circ\).

\[\begin{aligned}\tan15^\circ &= \frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}} \\ &= \frac{\sqrt3-1}{\sqrt3+1}\end{aligned}\]\[\begin{aligned}\text{L.H.S.} &= \cot15^\circ \\ &= \frac{\sqrt3+1}{\sqrt3-1} \\ &= \frac{(\sqrt3+1)^2}{3-1} \\ &= 2+\sqrt3 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.3 If \(\sin A=\frac35\) and \(\cos B=\frac{12}{13}\), find the value of:

  • a. \(\sin(A+B)\)
  • b. \(\cos(A+B)\)
  • c. \(\sin(A-B)\)
  • d. \(\cos(A-B)\)
  • e. \(\tan(A+B)\)
  • f. \(\cot(A-B)\)

(यदि \(\sin A=\frac35\) र \(\cos B=\frac{12}{13}\) भए मान पत्ता लगाउनुहोस् : )

a. \(\sin(A+B)\)

Here, \(\sin A=\frac35\) and \(\cos B=\frac{12}{13}\) (taking \(A\) and \(B\) acute).

\[\begin{aligned}\cos A &= \sqrt{1-\sin^2A} \\ &= \sqrt{1-\frac{9}{25}} \\ &= \frac45,\qquad \sin B \\ &= \sqrt{1-\cos^2B} \\ &= \sqrt{1-\frac{144}{169}} \\ &= \frac{5}{13}\end{aligned}\]\[\begin{aligned}\sin(A+B) &= \sin A\cos B+\cos A\sin B \\ &= \frac35\cdot\frac{12}{13}+\frac45\cdot\frac{5}{13} \\ &= \frac{36}{65}+\frac{20}{65} \\ &= \frac{56}{65}\end{aligned}\]
\(\sin(A+B)=\dfrac{56}{65}\)
b. \(\cos(A+B)\)

Here, \(\sin A=\frac35,\ \cos A=\frac45,\ \cos B=\frac{12}{13},\ \sin B=\frac{5}{13}\) (found in part (a)).

\[\begin{aligned}\cos(A+B) &= \cos A\cos B-\sin A\sin B \\ &= \frac45\cdot\frac{12}{13}-\frac35\cdot\frac{5}{13} \\ &= \frac{48}{65}-\frac{15}{65} \\ &= \frac{33}{65}\end{aligned}\]
\(\cos(A+B)=\dfrac{33}{65}\)
c. \(\sin(A-B)\)

Here, \(\sin A=\frac35,\ \cos A=\frac45,\ \cos B=\frac{12}{13},\ \sin B=\frac{5}{13}\) (found in part (a)).

\[\begin{aligned}\sin(A-B) &= \sin A\cos B-\cos A\sin B \\ &= \frac35\cdot\frac{12}{13}-\frac45\cdot\frac{5}{13} \\ &= \frac{36}{65}-\frac{20}{65} \\ &= \frac{16}{65}\end{aligned}\]
\(\sin(A-B)=\dfrac{16}{65}\)
d. \(\cos(A-B)\)

Here, \(\sin A=\frac35,\ \cos A=\frac45,\ \cos B=\frac{12}{13},\ \sin B=\frac{5}{13}\) (found in part (a)).

\[\begin{aligned}\cos(A-B) &= \cos A\cos B+\sin A\sin B \\ &= \frac45\cdot\frac{12}{13}+\frac35\cdot\frac{5}{13} \\ &= \frac{48}{65}+\frac{15}{65} \\ &= \frac{63}{65}\end{aligned}\]
\(\cos(A-B)=\dfrac{63}{65}\)
e. \(\tan(A+B)\)

Here, \(\sin A=\frac35,\ \cos A=\frac45,\ \cos B=\frac{12}{13},\ \sin B=\frac{5}{13}\) (found in part (a)).

Using parts (a) and (b):

\[\begin{aligned}\tan(A+B) &= \frac{\sin(A+B)}{\cos(A+B)} \\ &= \frac{56/65}{33/65} \\ &= \frac{56}{33}\end{aligned}\]
\(\tan(A+B)=\dfrac{56}{33}\)
f. \(\cot(A-B)\)

Here, \(\sin A=\frac35,\ \cos A=\frac45,\ \cos B=\frac{12}{13},\ \sin B=\frac{5}{13}\) (found in part (a)).

Using parts (c) and (d):

\[\begin{aligned}\cot(A-B) &= \frac{\cos(A-B)}{\sin(A-B)} \\ &= \frac{63/65}{16/65} \\ &= \frac{63}{16}\end{aligned}\]
\(\cot(A-B)=\dfrac{63}{16}\)

Q.4 Prove that:

  • a. \(\sin(A+B)+\sin(A-B)=2\sin A\cos B\)
  • b. \(\sin(A+B)-\sin(A-B)=2\cos A\sin B\)
  • c. \(\cos(A+B)+\cos(A-B)=2\cos A\cos B\)
  • d. \(\cos(A-B)-\cos(A+B)=2\sin A\sin B\)

(प्रमाणित गर्नुहोस् : )

a. \(\sin(A+B)+\sin(A-B)=2\sin A\cos B\)
\[\begin{aligned}\text{L.H.S.} &= \sin(A+B)+\sin(A-B) \\ &= (\sin A\cos B+\cos A\sin B)+(\sin A\cos B-\cos A\sin B) \\ &= 2\sin A\cos B \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\sin(A+B)-\sin(A-B)=2\cos A\sin B\)
\[\begin{aligned}\text{L.H.S.} &= \sin(A+B)-\sin(A-B) \\ &= (\sin A\cos B+\cos A\sin B)-(\sin A\cos B-\cos A\sin B) \\ &= 2\cos A\sin B \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\cos(A+B)+\cos(A-B)=2\cos A\cos B\)
\[\begin{aligned}\text{L.H.S.} &= \cos(A+B)+\cos(A-B) \\ &= (\cos A\cos B-\sin A\sin B)+(\cos A\cos B+\sin A\sin B) \\ &= 2\cos A\cos B \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\cos(A-B)-\cos(A+B)=2\sin A\sin B\)
\[\begin{aligned}\text{L.H.S.} &= \cos(A-B)-\cos(A+B) \\ &= (\cos A\cos B+\sin A\sin B)-(\cos A\cos B-\sin A\sin B) \\ &= 2\sin A\sin B \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.5 Prove that: (Without using calculator or table)

  • a. \(\sin75^\circ-\sin15^\circ=\frac{1}{\sqrt2}\)
  • b. \(\cos15^\circ-\sin75^\circ=0\)
  • c. \(\sin105^\circ-\cos75^\circ=\frac{1}{\sqrt2}\)
  • d. \(\cos105^\circ+\cos15^\circ=\frac{1}{\sqrt2}\)
  • e. \(\tan15^\circ+\cot15^\circ=4\)
  • f. \(\cos18^\circ-\sin18^\circ=\sqrt2\sin27^\circ\)
  • g. \(\sin35^\circ+\cos35^\circ=\sqrt2\cos10^\circ\)
  • h. \(\sqrt3\cos20^\circ+\sin20^\circ=2\sin80^\circ\)

(क्याल्कुलेटर वा तालिका प्रयोग नगरी सिद्ध गर्नुहोस् : )

a. \(\sin75^\circ-\sin15^\circ=\frac{1}{\sqrt2}\)

From Question 2, \(\sin75^\circ=\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}\) and \(\sin15^\circ=\cos75^\circ=\frac{\sqrt3-1}{2\sqrt2}\).

\[\begin{aligned}\text{L.H.S.} &= \frac{\sqrt3+1}{2\sqrt2}-\frac{\sqrt3-1}{2\sqrt2} \\ &= \frac{2}{2\sqrt2} \\ &= \frac{1}{\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\cos15^\circ-\sin75^\circ=0\)
\[\begin{aligned}\text{L.H.S.} &= \cos15^\circ-\sin75^\circ \\ &= \cos(90^\circ-75^\circ)-\sin75^\circ \\ &= \sin75^\circ-\sin75^\circ \\ &= 0 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\sin105^\circ-\cos75^\circ=\frac{1}{\sqrt2}\)

Here, \(\sin105^\circ=\sin(60^\circ+45^\circ)=\frac{\sqrt3+1}{2\sqrt2}\) and \(\cos75^\circ=\cos(45^\circ+30^\circ)=\frac{\sqrt3-1}{2\sqrt2}\).

\[\begin{aligned}\text{L.H.S.} &= \frac{\sqrt3+1}{2\sqrt2}-\frac{\sqrt3-1}{2\sqrt2} \\ &= \frac{2}{2\sqrt2} \\ &= \frac{1}{\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\cos105^\circ+\cos15^\circ=\frac{1}{\sqrt2}\)

From Question 2, \(\cos105^\circ=\frac{1-\sqrt3}{2\sqrt2}\) and \(\cos15^\circ=\frac{\sqrt3+1}{2\sqrt2}\).

\[\begin{aligned}\text{L.H.S.} &= \frac{1-\sqrt3}{2\sqrt2}+\frac{\sqrt3+1}{2\sqrt2} \\ &= \frac{2}{2\sqrt2} \\ &= \frac{1}{\sqrt2} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\tan15^\circ+\cot15^\circ=4\)

Here, \(\cot15^\circ=2+\sqrt3\) (Question 2f) and \(\tan15^\circ=\dfrac{1}{2+\sqrt3}=\dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}=2-\sqrt3\).

\[\begin{aligned}\text{L.H.S.} &= (2-\sqrt3)+(2+\sqrt3) \\ &= 4 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\cos18^\circ-\sin18^\circ=\sqrt2\sin27^\circ\)
\[\begin{aligned}\text{L.H.S.} &= \sqrt2\left(\frac{1}{\sqrt2}\cos18^\circ-\frac{1}{\sqrt2}\sin18^\circ\right) \\ &= \sqrt2(\sin45^\circ\cos18^\circ-\cos45^\circ\sin18^\circ) \\ &= \sqrt2\sin(45^\circ-18^\circ) \\ &= \sqrt2\sin27^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
g. \(\sin35^\circ+\cos35^\circ=\sqrt2\cos10^\circ\)
\[\begin{aligned}\text{L.H.S.} &= \sqrt2\left(\frac{1}{\sqrt2}\cos35^\circ+\frac{1}{\sqrt2}\sin35^\circ\right) \\ &= \sqrt2(\cos45^\circ\cos35^\circ+\sin45^\circ\sin35^\circ) \\ &= \sqrt2\cos(45^\circ-35^\circ) \\ &= \sqrt2\cos10^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
h. \(\sqrt3\cos20^\circ+\sin20^\circ=2\sin80^\circ\)
\[\begin{aligned}\text{L.H.S.} &= 2\left(\frac{\sqrt3}{2}\cos20^\circ+\frac12\sin20^\circ\right) \\ &= 2(\sin60^\circ\cos20^\circ+\cos60^\circ\sin20^\circ) \\ &= 2\sin(60^\circ+20^\circ) \\ &= 2\sin80^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.6 Prove that:

  • a. \(\sin(25^\circ+\alpha)\cos(65^\circ-\alpha)+\cos(25^\circ+\alpha)\sin(65^\circ-\alpha)=1\)
  • b. \(\cos(25^\circ+\alpha)\cos(65^\circ+\alpha)-\sin(25^\circ+\alpha)\sin(65^\circ+\alpha)=-\sin2\alpha\)
  • c. \(\cos(60^\circ+\beta)-\cos\beta+\cos(60^\circ-\beta)=0\)
  • d. \(\frac{\tan(45^\circ+\phi)-\tan\phi}{1+\tan(45^\circ+\phi)\tan\phi}=1\)

(प्रमाणित गर्नुहोस् : )

a. \(\sin(25^\circ+\alpha)\cos(65^\circ-\alpha)+\cos(25^\circ+\alpha)\sin(65^\circ-\alpha)=1\)

Let \(A=25^\circ+\alpha\) and \(B=65^\circ-\alpha\).

\[\begin{aligned}\text{L.H.S.} &= \sin A\cos B+\cos A\sin B \\ &= \sin(A+B) \\ &= \sin(25^\circ+\alpha+65^\circ-\alpha) \\ &= \sin90^\circ \\ &= 1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\cos(25^\circ+\alpha)\cos(65^\circ+\alpha)-\sin(25^\circ+\alpha)\sin(65^\circ+\alpha)=-\sin2\alpha\)

Let \(A=25^\circ+\alpha\) and \(B=65^\circ+\alpha\).

\[\begin{aligned}\text{L.H.S.} &= \cos A\cos B-\sin A\sin B \\ &= \cos(A+B) \\ &= \cos(90^\circ+2\alpha) \\ &= -\sin2\alpha \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\cos(60^\circ+\beta)-\cos\beta+\cos(60^\circ-\beta)=0\)

Using \(\cos(A+B)+\cos(A-B)=2\cos A\cos B\):

\[\begin{aligned}\text{L.H.S.} &= \cos(60^\circ+\beta)+\cos(60^\circ-\beta)-\cos\beta \\ &= 2\cos60^\circ\cos\beta-\cos\beta \\ &= 2\left(\frac12\right)\cos\beta-\cos\beta \\ &= 0 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\frac{\tan(45^\circ+\phi)-\tan\phi}{1+\tan(45^\circ+\phi)\tan\phi}=1\)

Using \(\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan(A-B)\) with \(A=45^\circ+\phi\) and \(B=\phi\):

\[\begin{aligned}\text{L.H.S.} &= \tan((45^\circ+\phi)-\phi) \\ &= \tan45^\circ \\ &= 1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.7 If \(A+B=\frac{\pi}{4}^c\), prove that:

  • a. \(\tan A+\tan B+\tan A\tan B=1\)
  • b. \((1+\tan A)(1+\tan B)=2\)

(यदि \(A+B=\frac{\pi}{4}^c\) भए प्रमाणित गर्नुहोस् : )

a. \(\tan A+\tan B+\tan A\tan B=1\)

Here, \(A+B=45^\circ\). Taking \(\tan\) on both sides:

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

That is, \(\tan A+\tan B+\tan A\tan B=1\).

Hence proved.
b. \((1+\tan A)(1+\tan B)=2\)

From part (a), \(\tan A+\tan B+\tan A\tan B=1\). Add \(1\) to both sides:

\[\begin{aligned}& 1+\tan A+\tan B+\tan A\tan B=2 \\ &\implies (1+\tan A)+\tan B(1+\tan A)=2 \\ &\implies (1+\tan A)(1+\tan B)=2\end{aligned}\]
Hence proved.
Q.8
a. If \(\tan X=\frac23\) and \(\tan Y=\frac15\), prove that \(X+Y=45^\circ\). (यदि \(\tan X=\frac23\) र \(\tan Y=\frac15\) भए \(X+Y=45^\circ\) सिद्ध गर्नुहोस् ।)

Here,

\[\begin{aligned}\tan(X+Y) &= \frac{\tan X+\tan Y}{1-\tan X\tan Y} \\ &= \frac{\frac23+\frac15}{1-\frac23\cdot\frac15} \\ &= \frac{13/15}{13/15} \\ &= 1\end{aligned}\]

Since \(\tan(X+Y)=1=\tan45^\circ\), we get \(X+Y=45^\circ\).

Hence proved.
b. If \(\tan A=m\) and \(\tan B=\frac1m\), prove that \(A+B=\frac{\pi}{2}^c\). (यदि \(\tan A=m\) र \(\tan B=\frac1m\) भए \(A+B=\frac{\pi}{2}^c\) सिद्ध गर्नुहोस् ।)

Here, \(\tan B=\frac1m\) gives \(\cot B=m\). So

\[\begin{aligned}& \tan A=m=\cot B=\tan\left(\frac{\pi}{2}-B\right) \\ &\implies A=\frac{\pi}{2}-B \\ &\implies A+B=\frac{\pi}{2}\end{aligned}\]
Hence proved.

Q.9 Prove that:

  • a. \(\frac{\cos35^\circ-\sin35^\circ}{\cos35^\circ+\sin35^\circ}=\tan10^\circ\)
  • b. \(\frac{\cos17^\circ+\sin17^\circ}{\cos17^\circ-\sin17^\circ}=\tan62^\circ\)
  • c. \(\frac{\cos8^\circ+\sin8^\circ}{\cos8^\circ-\sin8^\circ}=\cot37^\circ\)
  • d. \(\frac{\cos10^\circ-\sin10^\circ}{\cos10^\circ+\sin10^\circ}=\tan35^\circ\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\cos35^\circ-\sin35^\circ}{\cos35^\circ+\sin35^\circ}=\tan10^\circ\)

Divide the numerator and denominator by \(\cos35^\circ\):

\[\begin{aligned}\text{L.H.S.} &= \frac{1-\tan35^\circ}{1+\tan35^\circ} \\ &= \frac{\tan45^\circ-\tan35^\circ}{1+\tan45^\circ\tan35^\circ} \\ &= \tan(45^\circ-35^\circ) \\ &= \tan10^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{\cos17^\circ+\sin17^\circ}{\cos17^\circ-\sin17^\circ}=\tan62^\circ\)

Divide the numerator and denominator by \(\cos17^\circ\):

\[\begin{aligned}\text{L.H.S.} &= \frac{1+\tan17^\circ}{1-\tan17^\circ} \\ &= \frac{\tan45^\circ+\tan17^\circ}{1-\tan45^\circ\tan17^\circ} \\ &= \tan(45^\circ+17^\circ) \\ &= \tan62^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\frac{\cos8^\circ+\sin8^\circ}{\cos8^\circ-\sin8^\circ}=\cot37^\circ\)

Divide the numerator and denominator by \(\cos8^\circ\):

\[\begin{aligned}\text{L.H.S.} &= \frac{1+\tan8^\circ}{1-\tan8^\circ} \\ &= \tan(45^\circ+8^\circ) \\ &= \tan53^\circ \\ &= \tan(90^\circ-37^\circ) \\ &= \cot37^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\frac{\cos10^\circ-\sin10^\circ}{\cos10^\circ+\sin10^\circ}=\tan35^\circ\)

Divide the numerator and denominator by \(\cos10^\circ\):

\[\begin{aligned}\text{L.H.S.} &= \frac{1-\tan10^\circ}{1+\tan10^\circ} \\ &= \tan(45^\circ-10^\circ) \\ &= \tan35^\circ \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.10 Prove that:

  • a. \(\frac{\tan5A-\tan4A}{1+\tan5A\tan4A}=\tan A\)
  • b. \(\frac{\tan7A-\tan4A}{1+\tan7A\tan4A}=\tan3A\)
  • c. \(\frac{\tan(A+B)-\tan B}{1+\tan(A+B)\tan B}=\tan A\)
  • d. \(\tan(A+B)\tan(A-B)=\frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B}\)
  • e. \(\frac{\sin(A+B)}{\sin A\sin B}=\cot B+\cot A\)
  • f. \(\frac{\cos(A+B)}{\sin A\sin B}=\cot A\cot B-1\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\tan5A-\tan4A}{1+\tan5A\tan4A}=\tan A\)
\[\begin{aligned}\text{L.H.S.} &= \tan(5A-4A) \\ &= \tan A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{\tan7A-\tan4A}{1+\tan7A\tan4A}=\tan3A\)
\[\begin{aligned}\text{L.H.S.} &= \tan(7A-4A) \\ &= \tan3A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\frac{\tan(A+B)-\tan B}{1+\tan(A+B)\tan B}=\tan A\)
\[\begin{aligned}\text{L.H.S.} &= \tan((A+B)-B) \\ &= \tan A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\tan(A+B)\tan(A-B)=\frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B}\)
\[\begin{aligned}\text{L.H.S.} &= \left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)\left(\frac{\tan A-\tan B}{1+\tan A\tan B}\right) \\ &= \frac{\tan^2A-\tan^2B}{1-\tan^2A\tan^2B} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\frac{\sin(A+B)}{\sin A\sin B}=\cot B+\cot A\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin A\cos B+\cos A\sin B}{\sin A\sin B} \\ &= \frac{\sin A\cos B}{\sin A\sin B}+\frac{\cos A\sin B}{\sin A\sin B} \\ &= \cot B+\cot A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\frac{\cos(A+B)}{\sin A\sin B}=\cot A\cot B-1\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\cos A\cos B-\sin A\sin B}{\sin A\sin B} \\ &= \frac{\cos A\cos B}{\sin A\sin B}-1 \\ &= \cot A\cot B-1 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.11 Prove that:

  • a. \(\tan25^\circ+\tan20^\circ+\tan25^\circ\tan20^\circ=1\)
  • b. \(1-\tan27^\circ-\tan18^\circ=\tan27^\circ\tan18^\circ\)
  • c. \(\tan65^\circ-\tan20^\circ=1+\tan65^\circ\tan20^\circ\)
  • d. \(1+\cot18^\circ+\cot27^\circ=\cot18^\circ\cot27^\circ\)

(प्रमाणित गर्नुहोस् : )

a. \(\tan25^\circ+\tan20^\circ+\tan25^\circ\tan20^\circ=1\)

Since \(25^\circ+20^\circ=45^\circ\), take \(\tan\) on both sides:

\[\frac{\tan25^\circ+\tan20^\circ}{1-\tan25^\circ\tan20^\circ}=\tan45^\circ=1 \implies \tan25^\circ+\tan20^\circ=1-\tan25^\circ\tan20^\circ\]

Hence \(\tan25^\circ+\tan20^\circ+\tan25^\circ\tan20^\circ=1\).

Hence proved.
b. \(1-\tan27^\circ-\tan18^\circ=\tan27^\circ\tan18^\circ\)

Since \(27^\circ+18^\circ=45^\circ\):

\[\begin{aligned}& \tan(27^\circ+18^\circ)=1 \\ &\implies \tan27^\circ+\tan18^\circ=1-\tan27^\circ\tan18^\circ \\ &\implies 1-\tan27^\circ-\tan18^\circ=\tan27^\circ\tan18^\circ\end{aligned}\]
Hence proved.
c. \(\tan65^\circ-\tan20^\circ=1+\tan65^\circ\tan20^\circ\)

Since \(65^\circ-20^\circ=45^\circ\):

\[\begin{aligned}& \tan(65^\circ-20^\circ)=1 \\ &\implies \frac{\tan65^\circ-\tan20^\circ}{1+\tan65^\circ\tan20^\circ}=1 \\ &\implies \tan65^\circ-\tan20^\circ=1+\tan65^\circ\tan20^\circ\end{aligned}\]
Hence proved.
d. \(1+\cot18^\circ+\cot27^\circ=\cot18^\circ\cot27^\circ\)

Since \(18^\circ+27^\circ=45^\circ\), \(\cot(18^\circ+27^\circ)=1\):

\[\begin{aligned}& \frac{\cot18^\circ\cot27^\circ-1}{\cot18^\circ+\cot27^\circ}=1 \\ &\implies \cot18^\circ\cot27^\circ-1=\cot18^\circ+\cot27^\circ \\ &\implies 1+\cot18^\circ+\cot27^\circ=\cot18^\circ\cot27^\circ\end{aligned}\]
Hence proved.

Q.12 Prove that:

  • a. \(\tan50^\circ+\tan60^\circ+\tan70^\circ=\tan50^\circ\tan60^\circ\tan70^\circ\)
  • b. \(\tan10A-\tan6A-\tan4A=\tan10A\tan6A\tan4A\)
  • c. \(\tan20^\circ\tan30^\circ+\tan30^\circ\tan40^\circ+\tan40^\circ\tan20^\circ=1\)
  • d. \(\cot5A\cot4A-\cot9A\cot5A-\cot9A\cot4A=1\)

(प्रमाणित गर्नुहोस् : )

a. \(\tan50^\circ+\tan60^\circ+\tan70^\circ=\tan50^\circ\tan60^\circ\tan70^\circ\)

Since \(50^\circ+60^\circ+70^\circ=180^\circ\), we have \(50^\circ+60^\circ=180^\circ-70^\circ\).

\[\tan(50^\circ+60^\circ)=-\tan70^\circ \implies \frac{\tan50^\circ+\tan60^\circ}{1-\tan50^\circ\tan60^\circ}=-\tan70^\circ\]\[\tan50^\circ+\tan60^\circ=-\tan70^\circ+\tan50^\circ\tan60^\circ\tan70^\circ \implies \tan50^\circ+\tan60^\circ+\tan70^\circ=\tan50^\circ\tan60^\circ\tan70^\circ\]
Hence proved.
b. \(\tan10A-\tan6A-\tan4A=\tan10A\tan6A\tan4A\)

Since \(10A=6A+4A\):

\[\tan10A=\frac{\tan6A+\tan4A}{1-\tan6A\tan4A} \implies \tan10A-\tan10A\tan6A\tan4A=\tan6A+\tan4A\]\[\tan10A-\tan6A-\tan4A=\tan10A\tan6A\tan4A\]
Hence proved.
c. \(\tan20^\circ\tan30^\circ+\tan30^\circ\tan40^\circ+\tan40^\circ\tan20^\circ=1\)

Since \(20^\circ+40^\circ=90^\circ-30^\circ\):

\[\tan(20^\circ+40^\circ)=\cot30^\circ=\frac{1}{\tan30^\circ} \implies \frac{\tan20^\circ+\tan40^\circ}{1-\tan20^\circ\tan40^\circ}=\frac{1}{\tan30^\circ}\]\[\tan20^\circ\tan30^\circ+\tan30^\circ\tan40^\circ=1-\tan20^\circ\tan40^\circ\]

Hence \(\tan20^\circ\tan30^\circ+\tan30^\circ\tan40^\circ+\tan40^\circ\tan20^\circ=1\).

Hence proved.
d. \(\cot5A\cot4A-\cot9A\cot5A-\cot9A\cot4A=1\)

Since \(9A=5A+4A\):

\[\cot9A=\frac{\cot5A\cot4A-1}{\cot5A+\cot4A} \implies \cot9A\cot5A+\cot9A\cot4A=\cot5A\cot4A-1\]

Hence \(\cot5A\cot4A-\cot9A\cot5A-\cot9A\cot4A=1\).

Hence proved.

Q.13 Prove that:

  • a. \(\tan50^\circ-\tan40^\circ=2\tan10^\circ\)
  • b. \(\tan50^\circ=2\tan30^\circ+\tan20^\circ\)

(प्रमाणित गर्नुहोस् : )

a. \(\tan50^\circ-\tan40^\circ=2\tan10^\circ\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin50^\circ}{\cos50^\circ}-\frac{\sin40^\circ}{\cos40^\circ} \\ &= \frac{\sin50^\circ\cos40^\circ-\cos50^\circ\sin40^\circ}{\cos50^\circ\cos40^\circ} \\ &= \frac{\sin10^\circ}{\cos50^\circ\sin50^\circ} \\ &= \frac{2\sin10^\circ}{\sin100^\circ} \\ &= \frac{2\sin10^\circ}{\cos10^\circ} \\ &= 2\tan10^\circ \\ &= \text{R.H.S.}\end{aligned}\]

Here \(\cos40^\circ=\sin50^\circ\) and \(\sin100^\circ=\sin(90^\circ+10^\circ)=\cos10^\circ\).

Therefore, L.H.S. = R.H.S., proved.
b. \(\tan50^\circ=2\tan30^\circ+\tan20^\circ\)

Note on the printed question

As printed this relation is not true: \(\tan50^\circ\approx1.1918\) but \(2\tan30^\circ+\tan20^\circ\approx1.5187\). The relation that follows from part (a) is \(\tan50^\circ=\tan40^\circ+2\tan10^\circ\), which is probably what the textbook intends.

Proof of the corrected relation

From part (a), \(\tan50^\circ-\tan40^\circ=2\tan10^\circ\), so

\[\tan50^\circ=\tan40^\circ+2\tan10^\circ\]
\(\tan50^\circ=\tan40^\circ+2\tan10^\circ\) is proved.

Q.14 Prove that:

  • a. \(\frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)}=\frac{\tan\beta+\tan\gamma}{\tan\beta-\tan\gamma}\)
  • b. \(\frac{\sin(A+B)\sin(A-B)}{\cos^2A\cos^2B}=\tan^2A-\tan^2B\)
  • c. \(\frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}=\tan A\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\sin(\beta+\gamma)}{\sin(\beta-\gamma)}=\frac{\tan\beta+\tan\gamma}{\tan\beta-\tan\gamma}\)

Divide the numerator and denominator by \(\cos\beta\cos\gamma\):

\[\begin{aligned}\text{L.H.S.} &= \frac{\sin\beta\cos\gamma+\cos\beta\sin\gamma}{\sin\beta\cos\gamma-\cos\beta\sin\gamma} \\ &= \frac{\tan\beta+\tan\gamma}{\tan\beta-\tan\gamma} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\frac{\sin(A+B)\sin(A-B)}{\cos^2A\cos^2B}=\tan^2A-\tan^2B\)
\[\begin{aligned}\text{L.H.S.} &= \frac{\sin^2A-\sin^2B}{\cos^2A\cos^2B} \\ &= \frac{\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B}{\cos^2A\cos^2B} \\ &= \frac{\sin^2A\cos^2B-\cos^2A\sin^2B}{\cos^2A\cos^2B} \\ &= \tan^2A-\tan^2B \\ &= \text{R.H.S.}\end{aligned}\]

Here \(\sin(A+B)\sin(A-B)=\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B\).

Therefore, L.H.S. = R.H.S., proved.
c. \(\frac{\sin(A+B)+\sin(A-B)}{\cos(A+B)+\cos(A-B)}=\tan A\)

Using Question 4(a) and 4(c):

\[\begin{aligned}\text{L.H.S.} &= \frac{2\sin A\cos B}{2\cos A\cos B} \\ &= \frac{\sin A}{\cos A} \\ &= \tan A \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.15 Prove that:

  • a. \(\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}=0\)
  • b. \(\sin(A+B+C)=\cos A\cos B\cos C(\tan A+\tan B+\tan C-\tan A\tan B\tan C)\)
  • c. \(\cos(A+B+C)=\cos A\cos B\cos C(1-\tan A\tan B-\tan B\tan C-\tan C\tan A)\)
  • d. \(\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\)

(प्रमाणित गर्नुहोस् : )

a. \(\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}=0\)

Here, \(\dfrac{\sin(A-B)}{\cos A\cos B}=\dfrac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}=\tan A-\tan B\). Similarly the other two terms are \(\tan B-\tan C\) and \(\tan C-\tan A\).

\[\begin{aligned}\text{L.H.S.} &= (\tan A-\tan B)+(\tan B-\tan C)+(\tan C-\tan A) \\ &= 0 \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\sin(A+B+C)=\cos A\cos B\cos C(\tan A+\tan B+\tan C-\tan A\tan B\tan C)\)

Here,

\[\begin{aligned}\sin(A+B+C) &= \sin(A+B)\cos C+\cos(A+B)\sin C \\ &= (\sin A\cos B+\cos A\sin B)\cos C+(\cos A\cos B-\sin A\sin B)\sin C \\ &= \sin A\cos B\cos C+\cos A\sin B\cos C+\cos A\cos B\sin C-\sin A\sin B\sin C \\ &= \cos A\cos B\cos C\left(\tan A+\tan B+\tan C-\tan A\tan B\tan C\right)\end{aligned}\]

(taking \(\cos A\cos B\cos C\) common).

Hence proved.
c. \(\cos(A+B+C)=\cos A\cos B\cos C(1-\tan A\tan B-\tan B\tan C-\tan C\tan A)\)

Here,

\[\begin{aligned}\cos(A+B+C) &= \cos(A+B)\cos C-\sin(A+B)\sin C \\ &= (\cos A\cos B-\sin A\sin B)\cos C-(\sin A\cos B+\cos A\sin B)\sin C \\ &= \cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C \\ &= \cos A\cos B\cos C(1-\tan A\tan B-\tan C\tan A-\tan B\tan C)\end{aligned}\]

(taking \(\cos A\cos B\cos C\) common).

Hence proved.
d. \(\tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\)

Using parts (b) and (c):

\[\begin{aligned}\text{L.H.S.} &= \tan(A+B+C) \\ &= \frac{\sin(A+B+C)}{\cos(A+B+C)} \\ &= \frac{\cos A\cos B\cos C(\tan A+\tan B+\tan C-\tan A\tan B\tan C)}{\cos A\cos B\cos C(1-\tan A\tan B-\tan B\tan C-\tan C\tan A)} \\ &= \frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A} \\ &= \text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
Answer
1. A. b   B. c   C. b   D. a   E. d
3. \(\sin(A+B)=\frac{56}{65}\), \(\cos(A+B)=\frac{33}{65}\), \(\sin(A-B)=\frac{16}{65}\), \(\cos(A-B)=\frac{63}{65}\), \(\tan(A+B)=\frac{56}{33}\), \(\cot(A-B)=\frac{63}{16}\)
2, 4–15. Proved (see the note for Question 13b)

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

What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.1 Compound Angles? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.1 Compound Angles मा के के समावेश छ ?)

This page solves all 64 questions of Exercise 7.1 Compound Angles, in textbook order, with the Nepali question in brackets and an Answer box at the end. Compound angles examples, compound angle formula with proof and cot formula. All Exercise 7.1 questions solved for Class 10 Optional Math SEE Nepal.

What are compound angles in trigonometry? (त्रिकोणमितिमा संयुक्त कोण भनेको के हो ?)

A compound angle is an angle written as the sum or difference of two angles, such as \(A+B\) or \(A-B\). Exercise 7.1 uses the compound angle formulas to find exact values, to find \(\sin(A+B)\) from given ratios and to prove identities.

What is the compound angle formula in trigonometry? (संयुक्त कोणका सूत्रहरू के के हुन् ?)
  • \(\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\)
  • \(\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\)
  • \(\tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\)
What is the compound angle formula for cot? (\(\cot(A\pm B)\) को सूत्र के हो ?)

\(\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}\) and \(\cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}\). They follow from the tangent formulas by taking reciprocals (used in Exercise 7.1, Question 11d and 12d).

What is the proof of the compound angle formula? (संयुक्त कोणको सूत्रको प्रमाण कसरी गर्ने ?)

Take two points \(P(\cos A,\sin A)\) and \(Q(\cos B,\sin B)\) on the unit circle. The distance formula and the cosine rule both give \(PQ^2\); equating them proves \(\cos(A-B)=\cos A\cos B+\sin A\sin B\). Replace \(B\) by \(-B\) to get \(\cos(A+B)\), and use \(\sin x=\cos(90^\circ-x)\) to get the sine formulas.

Can you show compound angles examples? (संयुक्त कोणका उदाहरणहरू देखाउनुहोस् ।)

Write the angle as a sum or difference of standard angles, then apply the formula. Example: \(\sin75^\circ=\sin(45^\circ+30^\circ)=\dfrac{\sqrt3+1}{2\sqrt2}\) (Exercise 7.1, Question 2b). Another example: \(\cos15^\circ=\cos(45^\circ-30^\circ)=\dfrac{\sqrt3+1}{2\sqrt2}\) (Question 2a).

Where can I find compound angles trigonometry questions and problems with answers? (संयुक्त कोणका प्रश्न र समाधान कहाँ पाइन्छ ?)

This page solves all the questions of Exercise 7.1: five multiple choice questions and fifteen written questions with their parts, each with a worked solution and an Answer box at the end. Question 2 gives exact values, Question 3 finds values from given ratios, and Questions 4 to 15 are identities and proofs.

What is a compound angle in woodworking? Is it the same as in trigonometry? (काठको काममा संयुक्त कोण भनेको के हो ? के यो त्रिकोणमितिको जस्तै हो ?)

No. In woodworking a compound angle is a cut made with two angles at once, a miter angle and a bevel angle. In Class 10 trigonometry a compound angle means an angle of the form \(A\pm B\), and the formulas above give its sine, cosine and tangent.

How do I find the exact value of \(\sin75^\circ\) or \(\cos15^\circ\)? (\(\sin75^\circ\) वा \(\cos15^\circ\) को ठ्याक्कै मान कसरी निकाल्ने ?)

Write the angle as a sum or difference of standard angles, such as \(75^\circ=45^\circ+30^\circ\), then use the formula. For example \(\sin75^\circ=\dfrac{\sqrt3+1}{2\sqrt2}\) (Question 2b).

How do I find \(\sin(A+B)\) when only \(\sin A\) and \(\cos B\) are given? (केवल \(\sin A\) र \(\cos B\) दिइएको छ भने \(\sin(A+B)\) कसरी निकाल्ने ?)

Find \(\cos A=\sqrt{1-\sin^2A}\) and \(\sin B=\sqrt{1-\cos^2B}\), then substitute in the formula. See Question 3, where \(\sin(A+B)=\dfrac{56}{65}\).

How do I prove identities like \(\tan A+\tan B+\tan A\tan B=1\)? (\(\tan A+\tan B+\tan A\tan B=1\) जस्ता सर्वसमिका कसरी प्रमाणित गर्ने ?)

If \(A+B=45^\circ\), take \(\tan\) of both sides: \(\dfrac{\tan A+\tan B}{1-\tan A\tan B}=1\), then rearrange (Question 7).

How do I use \(45^\circ\) to simplify \(\dfrac{\cos35^\circ-\sin35^\circ}{\cos35^\circ+\sin35^\circ}\)? (\(45^\circ\) प्रयोग गरी यस्ता भिन्न कसरी सरल गर्ने ?)

Divide the numerator and denominator by \(\cos35^\circ\) to get \(\dfrac{1-\tan35^\circ}{1+\tan35^\circ}\). Since \(\tan45^\circ=1\), this is \(\tan(45^\circ-35^\circ)=\tan10^\circ\) (Question 9).

What is the identity for \(\tan(A+B+C)\)? (\(\tan(A+B+C)\) को सर्वसमिका के हो ?)

\(\tan(A+B+C)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\). It follows by dividing the sine and cosine expansions of Question 15.

Is every printed identity in Exercise 7.1 true? (के अभ्यास ७.१ मा छापिएका सबै सर्वसमिका सही छन् ?)

Not always. In Question 13(b) as printed, \(\tan50^\circ=2\tan30^\circ+\tan20^\circ\) is false (the relation that follows from part (a) is \(\tan50^\circ=\tan40^\circ+2\tan10^\circ\)). A quick numerical check with a calculator is a good habit.

Are these Exercise 7.1 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)

Yes, for students who study Optional Mathematics. Exercise 7.1 builds the compound angle skills used in all later exercises of Unit 7.

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