Conditional Identity in Trigonometry: Class 10 Optional Mathematics Exercise 7.5
Conditional identity formula, examples and conditional trigonometric identities class 10
25 solved questions with English and Nepali questions, textbook-style step by step solutions, key formulas and frequently asked questions for the Nepal Grade 10 SEE.
1. Conditional Identities: Class 10 Optional Mathematics Trigonometry Overview and Key Formulas
Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.5 Conditional Identities
A conditional identity is a relation that is true only when a given condition holds, such as \(A+B+C=180^\circ\). This page gives the conditional identity formulas for the angles of a triangle, conditional identity examples and step by step solutions to every question of Exercise 7.5, the conditional trigonometric identities of 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.
- Conditional identities (Exercise 7.5): an identity such as \(\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C\) is true only when a condition such as \(A+B+C=180^\circ\) holds. Use \(A+B=180^\circ-C\), so \(\sin(A+B)=\sin C\) and \(\cos(A+B)=-\cos C\).
Key formulas to remember
For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about trigonometric identities for extra understanding.
2. Exercise 7.5 Conditional Identities: Class 10 Optional Mathematics Trigonometry
Exercise 7.5 Conditional Identities: 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 statements is correct? (तलका मध्ये कुन कथन सही छ ?)
- a) Conditions are necessary in identities.
- b) Conditional identities are true in all situations.
- c) Identities are always true.
- d) Conditional identities do not need to be verified.
Correct Answer: c) Identities are always true.
Solution:
- An identity is true for all values of the angles for which both sides are defined.
- A conditional identity is true only when the given condition holds, for example \(A+B+C=180^\circ\).
(सर्वसमिका सधैं सत्य हुन्छ । अनुबन्धित सर्वसमिका दिइएको सर्त पूरा हुँदा मात्र सत्य हुन्छ ।)
B. If \(\tan A=1\), what is the value of \(A\)? (\(\tan A=1\) भएमा \(A\) को मान कति हुन्छ ?)
- a) \(0^\circ\)
- b) \(30^\circ\)
- c) \(45^\circ\)
- d) \(60^\circ\)
Correct Answer: c) \(45^\circ\)
Solution:
- Since \(\tan45^\circ=1\), we get \(A=45^\circ\).
(\(\tan45^\circ=1\) भएकाले \(A=45^\circ\) हुन्छ ।)
C. Under what condition is \(\sin A=\cos B\) true? (\(\sin A=\cos B\) कुन अवस्थामा सही हुन्छ ?)
- a) \(A=B\)
- b) \(A+B=90^\circ\)
- c) \(A-B=90^\circ\)
- d) \(A+B=180^\circ\)
Correct Answer: b) \(A+B=90^\circ\)
Solution:
- Since \(\cos B=\sin(90^\circ-B)\), the relation \(\sin A=\sin(90^\circ-B)\) gives \(A=90^\circ-B\).
(\(\cos B=\sin(90^\circ-B)\) भएकाले \(A+B=90^\circ\) हुन्छ ।)
D. Which of the following is a conditional identity? (तलका मध्ये कुन अनुबन्धित सर्वसमिका हो ?)
- a) \(\sin^2A+\cos^2A=1\)
- b) \(1+\tan^2A=\sec^2A\)
- c) If \(A+B=90^\circ\) then \(\cot A=\tan B\)
- d) \(\sin2A=2\sin A\cos A\)
Correct Answer: c) If \(A+B=90^\circ\) then \(\cot A=\tan B\)
Solution:
- Options (a), (b) and (d) are true for all angles for which they are defined.
- Option (c) is true only under the condition \(A+B=90^\circ\), so it is a conditional identity.
(विकल्प (ग) \(A+B=90^\circ\) सर्त पूरा हुँदा मात्र सत्य हुन्छ, त्यसैले यो अनुबन्धित सर्वसमिका हो ।)
E. If \(A+B+C=180^\circ\), which one is correct? (यदि \(A+B+C=180^\circ\) हुँदा कुन सही हुन्छ ?)
- a) \(\sin(A+B)=\cos C\)
- b) \(\sin(A+B)=\sin C\)
- c) \(\sin(A+B)=-\sin C\)
- d) \(\sin(A+B)=-\cos C\)
Correct Answer: b) \(\sin(A+B)=\sin C\)
Solution:
- Here \(A+B=180^\circ-C\).
(\(A+B=180^\circ-C\) भएकाले \(\sin(A+B)=\sin C\) हुन्छ ।)
Exercise 7.5 Conditional Identities: Written Questions (लेख्ने प्रश्न)
Q.2 The three angles of a triangle are \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\). Verify that:
- a. \(\sin(\alpha+\beta)=\sin\gamma\)
- b. \(\cos(\beta+\gamma)=-\cos\alpha\)
- c. \(\tan(\alpha+\beta)=-\tan\gamma\)
- d. \(\cot(\alpha+\gamma)=-\cot\beta\)
- e. \(\sin\left(\dfrac\alpha2+\dfrac\beta2\right)=\cos\dfrac\gamma2\)
- f. \(\cos\left(\dfrac\alpha2+\dfrac\beta2\right)=\sin\dfrac\gamma2\)
(त्रिभुजका तीनओटा कोण \(\alpha=30^\circ\), \(\beta=90^\circ\) र \(\gamma=60^\circ\) भए परीक्षण गर्नुहोस् : )
Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\sin(30^\circ+90^\circ)=\sin120^\circ=\frac{\sqrt3}{2} \\ \text{R.H.S.}&=\sin60^\circ=\frac{\sqrt3}{2} \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\cos(90^\circ+60^\circ)=\cos150^\circ=-\frac{\sqrt3}{2} \\ \text{R.H.S.}&=-\cos30^\circ=-\frac{\sqrt3}{2} \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\tan(30^\circ+90^\circ)=\tan120^\circ=-\sqrt3 \\ \text{R.H.S.}&=-\tan60^\circ=-\sqrt3 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\cot(30^\circ+60^\circ)=\cot90^\circ=0 \\ \text{R.H.S.}&=-\cot90^\circ=0 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\sin(15^\circ+45^\circ)=\sin60^\circ=\frac{\sqrt3}{2} \\ \text{R.H.S.}&=\cos30^\circ=\frac{\sqrt3}{2} \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=30^\circ\), \(\beta=90^\circ\) and \(\gamma=60^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\cos(15^\circ+45^\circ)=\cos60^\circ=\frac12 \\ \text{R.H.S.}&=\sin30^\circ=\frac12 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Q.3 If \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), verify that:
- a. \(\tan A+\tan B+\tan C=\tan A\tan B\tan C\)
- b. \(\tan2A+\tan2B+\tan2C=\tan2A\tan2B\tan2C\)
- c. \(\cot A\cot B+\cot B\cot C+\cot C\cot A=1\)
- d. \(\cot2A\cot2B+\cot2B\cot2C+\cot2C\cot2A=1\)
- e. \(\cot3A\cot3B+\cot3B\cot3C+\cot3C\cot3A=0\)
(यदि \(A=30^\circ\), \(B=120^\circ\) र \(C=30^\circ\) भए परीक्षण गर्नुहोस् : )
Given \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), so \(A+B+C=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\frac1{\sqrt3}-\sqrt3+\frac1{\sqrt3}=\frac{2-3}{\sqrt3}=-\frac1{\sqrt3} \\ \text{R.H.S.}&=\frac1{\sqrt3}\cdot(-\sqrt3)\cdot\frac1{\sqrt3}=-\frac1{\sqrt3} \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), so \(A+B+C=180^\circ\).
Here \(2A=60^\circ\), \(2B=240^\circ\), \(2C=60^\circ\) and \(\tan240^\circ=\tan(180^\circ+60^\circ)=\sqrt3\).
\[\begin{aligned}\text{L.H.S.}&=\tan60^\circ+\tan240^\circ+\tan60^\circ=\sqrt3+\sqrt3+\sqrt3=3\sqrt3 \\ \text{R.H.S.}&=\sqrt3\cdot\sqrt3\cdot\sqrt3=3\sqrt3 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), so \(A+B+C=180^\circ\).
Here \(\cot30^\circ=\sqrt3\) and \(\cot120^\circ=-\frac1{\sqrt3}\).
\[\begin{aligned}\text{L.H.S.}&=\sqrt3\left(-\frac1{\sqrt3}\right)+\left(-\frac1{\sqrt3}\right)\sqrt3+\sqrt3\cdot\sqrt3 \\ &=-1-1+3=1 \\ &=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), so \(A+B+C=180^\circ\).
Here \(2A=60^\circ\), \(2B=240^\circ\), \(2C=60^\circ\) and \(\cot60^\circ=\cot240^\circ=\frac1{\sqrt3}\).
\[\begin{aligned}\text{L.H.S.}&=\frac13+\frac13+\frac13=1 \\ &=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=120^\circ\) and \(C=30^\circ\), so \(A+B+C=180^\circ\).
Note on the printed question
Here \(3A=90^\circ\), \(3B=360^\circ\) and \(3C=90^\circ\). But \(\cot360^\circ=\dfrac{\cos360^\circ}{\sin360^\circ}\) is not defined, so the expression cannot be evaluated for these values, and the relation cannot be verified as printed.
Where it is defined, this expression equals \(1\), not \(0\). Put \(X=3A\), \(Y=3B\), \(Z=3C\), so \(X+Y+Z=540^\circ\):
\[\begin{aligned}\cot(X+Y)&=\cot(540^\circ-Z)=\cot(180^\circ-Z)=-\cot Z \\ \frac{\cot X\cot Y-1}{\cot X+\cot Y}&=-\cot Z \\ \cot X\cot Y+\cot Y\cot Z+\cot Z\cot X&=1\end{aligned}\]Check with \(A=B=45^\circ\), \(C=90^\circ\): \(\cot135^\circ=-1\), \(\cot135^\circ=-1\), \(\cot270^\circ=0\), so the sum is \((-1)(-1)+(-1)(0)+(0)(-1)=1\).
Q.4 If the three angles of \(\Delta ABC\) are \(A=30^\circ\), \(B=90^\circ\) and \(C=60^\circ\), verify that:
- a. \(\sin2A-\sin2B+\sin2C=4\cos A\sin B\cos C\)
- b. \(\sin2A-\sin2B-\sin2C=-4\sin A\cos B\cos C\)
- c. \(\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C\)
- d. \(\cos2A-\cos2B+\cos2C=1-4\sin A\cos B\sin C\)
(यदि \(\Delta ABC\) का तीनओटा कोण \(A=30^\circ\), \(B=90^\circ\) र \(C=60^\circ\) भए परीक्षण गर्नुहोस् : )
Given \(A=30^\circ\), \(B=90^\circ\) and \(C=60^\circ\), so \(A+B+C=180^\circ\). Then \(2A=60^\circ\), \(2B=180^\circ\), \(2C=120^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\sin60^\circ-\sin180^\circ+\sin120^\circ=\frac{\sqrt3}{2}-0+\frac{\sqrt3}{2}=\sqrt3 \\ \text{R.H.S.}&=4\cos30^\circ\sin90^\circ\cos60^\circ=4\cdot\frac{\sqrt3}{2}\cdot1\cdot\frac12=\sqrt3 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=90^\circ\) and \(C=60^\circ\), so \(A+B+C=180^\circ\). Then \(2A=60^\circ\), \(2B=180^\circ\), \(2C=120^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\frac{\sqrt3}{2}-0-\frac{\sqrt3}{2}=0 \\ \text{R.H.S.}&=-4\sin30^\circ\cos90^\circ\cos60^\circ=-4\cdot\frac12\cdot0\cdot\frac12=0 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=90^\circ\) and \(C=60^\circ\), so \(A+B+C=180^\circ\). Then \(2A=60^\circ\), \(2B=180^\circ\), \(2C=120^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\cos60^\circ+\cos180^\circ+\cos120^\circ=\frac12-1-\frac12=-1 \\ \text{R.H.S.}&=-1-4\cos30^\circ\cos90^\circ\cos60^\circ=-1-0=-1 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(A=30^\circ\), \(B=90^\circ\) and \(C=60^\circ\), so \(A+B+C=180^\circ\). Then \(2A=60^\circ\), \(2B=180^\circ\), \(2C=120^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\cos60^\circ-\cos180^\circ+\cos120^\circ=\frac12+1-\frac12=1 \\ \text{R.H.S.}&=1-4\sin30^\circ\cos90^\circ\sin60^\circ=1-0=1 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Q.5 The three angles of a triangle are \(\alpha=45^\circ\), \(\beta=45^\circ\) and \(\gamma=90^\circ\). Verify that:
- a. \(\sin^2\alpha+\sin^2\beta+\sin^2\gamma=4\sin\alpha\sin\beta\sin\gamma\)
- b. \(\sin^2\alpha+\sin^2\beta-\sin^2\gamma=1-2\cos\alpha\cos\beta\sin\gamma\)
- c. \(\cos^2\alpha-\cos^2\beta+\cos^2\gamma=1-2\sin\alpha\cos\beta\sin\gamma\)
- d. \(\cos^2\alpha-\cos^2\beta-\cos^2\gamma=2\cos\alpha\sin\beta\sin\gamma-1\)
(त्रिभुजका तीनओटा कोण \(\alpha=45^\circ\), \(\beta=45^\circ\) र \(\gamma=90^\circ\) भए परीक्षण गर्नुहोस् : )
Given \(\alpha=45^\circ\), \(\beta=45^\circ\) and \(\gamma=90^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\sin^245^\circ+\sin^245^\circ+\sin^290^\circ=\frac12+\frac12+1=2 \\ \text{R.H.S.}&=4\sin45^\circ\sin45^\circ\sin90^\circ=4\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}\cdot1=2 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Note on the printed question
The two sides agree for these particular angles only because \(\cos\gamma=\cos90^\circ=0\). The relation is not an identity: the identity for \(\alpha+\beta+\gamma=180^\circ\) is \(\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2+2\cos\alpha\cos\beta\cos\gamma\). For \(30^\circ,60^\circ,90^\circ\) the printed form gives \(2\) on the left and \(4\cdot\frac12\cdot\frac{\sqrt3}{2}\cdot1=\sqrt3\) on the right.
Given \(\alpha=45^\circ\), \(\beta=45^\circ\) and \(\gamma=90^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\frac12+\frac12-1=0 \\ \text{R.H.S.}&=1-2\cos45^\circ\cos45^\circ\sin90^\circ=1-2\cdot\frac12\cdot1=0 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Note on the printed question
This holds for these angles, but it is not an identity. The identity for \(\alpha+\beta+\gamma=180^\circ\) is \(\sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma\). For \(30^\circ,60^\circ,90^\circ\) the left side is \(0\) but \(1-2\cos30^\circ\cos60^\circ\sin90^\circ\approx0.134\).
Given \(\alpha=45^\circ\), \(\beta=45^\circ\) and \(\gamma=90^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\frac12-\frac12+0=0 \\ \text{R.H.S.}&=1-2\sin45^\circ\cos45^\circ\sin90^\circ=1-2\cdot\frac12\cdot1=0 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Given \(\alpha=45^\circ\), \(\beta=45^\circ\) and \(\gamma=90^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).
\[\begin{aligned}\text{L.H.S.}&=\frac12-\frac12-0=0 \\ \text{R.H.S.}&=2\cos45^\circ\sin45^\circ\sin90^\circ-1=2\cdot\frac12\cdot1-1=0 \\ \text{L.H.S.}&=\text{R.H.S.}\end{aligned}\]Q.6 Find out for which of the following conditions the identity \(\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C\) is valid.
(सर्वसमिका \(\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C\) तल दिइएका मध्ये कुन अवस्थामा मात्र मान्य हुन्छ, पत्ता लगाउनुहोस् ।)
- a) \(A+B=90^\circ\) and \(C=90^\circ\)
- b) \(A+B+C=180^\circ\)
- c) \(A+B+C=360^\circ\)
- d) \(A+B+C=90^\circ\)
Let \(A+B+C=180^\circ\), so \(A+B=180^\circ-C\). Then \(\sin(A+B)=\sin C\) and \(\cos C=-\cos(A+B)\).
\[\begin{aligned}\text{L.H.S.}&=2\sin(A+B)\cos(A-B)+\sin2C \\ &=2\sin C\cos(A-B)+2\sin C\cos C \\ &=2\sin C\left[\cos(A-B)+\cos C\right] \\ &=2\sin C\left[\cos(A-B)-\cos(A+B)\right] \\ &=2\sin C\cdot2\sin A\sin B \\ &=4\sin A\sin B\sin C \\ &=\text{R.H.S.}\end{aligned}\]So the identity is valid when \(A+B+C=180^\circ\). (Option (a) is a special case of this condition, since \(A+B=90^\circ\) and \(C=90^\circ\) also give \(A+B+C=180^\circ\).)
1. A. c B. c C. b D. c E. b
2, 3(a)–(d), 4, 5. Verified (see the notes for Questions 5a and 5b)
3(e). Not defined at \(B=120^\circ\) (see the note); the sum equals \(1\) where defined
6. b. \(A+B+C=180^\circ\)
3. Frequently Asked Questions on Conditional Identity in Trigonometry (Class 10 Optional Mathematics)
What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.5 Conditional Identities? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.5 Conditional Identities मा के के समावेश छ ?)
This page solves all 25 questions of Exercise 7.5 Conditional Identities, in textbook order, with the Nepali question in brackets and an Answer box at the end. Conditional identity formula and examples in trigonometry with all Exercise 7.5 questions solved. Conditional trigonometric identities for Class 10 SEE.
What is a conditional identity in trigonometry? (त्रिकोणमितिमा अनुबन्धित सर्वसमिका भनेको के हो ?)
A conditional identity is a relation that is true only when a given condition holds. For example \(\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C\) is true when \(A+B+C=180^\circ\) (Question 6), whereas an ordinary identity such as \(\sin^2A+\cos^2A=1\) is true for all angles.
What is the conditional identity formula for a triangle? (त्रिभुजका कोणका लागि अनुबन्धित सर्वसमिकाका सूत्रहरू के के हुन् ?)
If \(A+B+C=180^\circ\):
- \(\tan A+\tan B+\tan C=\tan A\tan B\tan C\)
- \(\cot A\cot B+\cot B\cot C+\cot C\cot A=1\)
- \(\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C\)
- \(\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C\)
Can you give a conditional identity example? (अनुबन्धित सर्वसमिकाको उदाहरण दिनुहोस् ।)
Take \(\alpha=30^\circ\), \(\beta=90^\circ\), \(\gamma=60^\circ\) (sum \(180^\circ\)). Then \(\sin(\alpha+\beta)=\sin120^\circ=\dfrac{\sqrt3}{2}=\sin\gamma\), which verifies the conditional identity (Question 2a).
How do I use the condition \(A+B+C=180^\circ\) in a proof? (\(A+B+C=180^\circ\) सर्त प्रमाणमा कसरी प्रयोग गर्ने ?)
Write \(A+B=180^\circ-C\). Then \(\sin(A+B)=\sin C\), \(\cos(A+B)=-\cos C\) and \(\tan(A+B)=-\tan C\). Combine the first two terms with a sum to product formula and use these relations to simplify (Question 6).
Is a conditional identity in trigonometry the same as conditional identity in logic or discrete math? (के त्रिकोणमितिको अनुबन्धित सर्वसमिका तर्कशास्त्र वा डिस्क्रिट म्याथको जस्तै हो ?)
No. In logic and discrete math a conditional means “if p then q”. In trigonometry a conditional identity is an equation between trigonometric expressions that holds only under a condition on the angles, for example that they add up to \(180^\circ\).
Where can I find conditional trigonometric identities class 10 questions with answers? (कक्षा १० का अनुबन्धित त्रिकोणमितीय सर्वसमिकाका प्रश्न र समाधान कहाँ पाइन्छ ?)
This page solves all the questions of Exercise 7.5, five multiple choice questions and the written verification questions, each with a worked solution and an Answer box at the end.
Are there misprints in Exercise 7.5? (के अभ्यास ७.५ मा छापाइका गल्ती छन् ?)
Yes. In Question 3(e) the expression is not defined at \(B=120^\circ\). In Question 5(a) and 5(b) the printed relations hold for the given angles but are not identities; the solutions give the correct identities.
Are these Exercise 7.5 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)
Yes. Exercise 7.5 trains students to verify and prove conditional identities, which is a common SEE-style question type in Optional Mathematics.
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