CBSE Class 10 Mathematics Trigonometry Assignment Set 24

Trigonometric Identities (Exercise III)

Question 1. Express the trigonometric ratios \( \sin A,\ \sec A \) and \( \tan A \) in terms of \( \cot A \).
Answer:
Using \( \operatorname{cosec}^2 A - \cot^2 A = 1 \): \( \frac{1}{\sin^2 A} = 1 + \cot^2 A \Rightarrow \sin^2 A = \frac{1}{1 + \cot^2 A} \).
Since A is acute, \( \sin A = \frac{1}{\sqrt{1 + \cot^2 A}} \).
\( \sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A} \Rightarrow \sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A} \).
\( \tan A = \frac{1}{\cot A} \).
In simple words: Use the identity that links cosec² A with cot² A, then take positive square roots for an acute angle.

Teacher's Note:
1. Remember the identity cosec² A = 1 + cot² A.
2. For an acute angle every ratio is positive, so choose the positive root.

Question 2. Write all the other trigonometric ratios of \( \angle A \) in terms of \( \sec A \).
Answer:
\( \cos A = \frac{1}{\sec A} \).
\( \sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A} \Rightarrow \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \).
\( \operatorname{cosec} A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}} \).
\( \tan^2 A = \sec^2 A - 1 \Rightarrow \tan A = \sqrt{\sec^2 A - 1} \) and \( \cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}} \).
In simple words: Every ratio can be written using sec A with the help of the identities 1 + tan² A = sec² A and sin² A + cos² A = 1.

Teacher's Note:
1. Take the positive square root because the angle is acute.
2. Write cos A first; the other ratios follow from it.

Question 3. Choose the correct option and justify your choice. (i) \( 9\sec^2 A - 9\tan^2 A = \) (a) 1 (b) 9 (c) 8 (d) 0 (ii) \( (1 + \tan \theta + \sec \theta)(1 + \cot \theta - \operatorname{cosec} \theta) = \) (a) 0 (b) 1 (c) 2 (d) \( -1 \) (iii) \( (\sec A + \tan A)(1 - \sin A) = \) (a) \( \sec A \) (b) \( \sin A \) (c) \( \operatorname{cosec} A \) (d) \( \cos A \) (iv) \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \) (a) \( \sec^2 A \) (b) \( -1 \) (c) \( \cot^2 A \) (d) \( \tan^2 A \)
Answer:
(i) Option (b). \( 9\sec^2 A - 9\tan^2 A = 9(\sec^2 A - \tan^2 A) = 9 \times 1 = 9 \).
(ii) Option (c). \( (1 + \tan \theta + \sec \theta)(1 + \cot \theta - \operatorname{cosec} \theta) = \left( 1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} \right)\left( 1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \right) = \frac{(\cos \theta + \sin \theta + 1)(\sin \theta + \cos \theta - 1)}{\cos \theta \sin \theta} \).
\( = \frac{(\sin \theta + \cos \theta)^2 - 1}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2 \).
(iii) Option (d). \( (\sec A + \tan A)(1 - \sin A) = \left( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \right)(1 - \sin A) = \frac{(1 + \sin A)(1 - \sin A)}{\cos A} = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \).
(iv) Option (d). \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \).
In simple words: Reduce each expression with the identities; the results are 9, 2, cos A and tan² A.

Teacher's Note:
1. Look for the identities sec² A − tan² A = 1 and cosec² A − cot² A = 1.
2. Use (a + b)(a − b) = a² − b² wherever the brackets allow it.

Question 4(i). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( (\operatorname{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \)
Answer:
LHS \( = (\operatorname{cosec} \theta - \cot \theta)^2 = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \).
\( = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(ii). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2\sec A \)
Answer:
LHS \( = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1 + \sin A)\cos A} \).
Since \( \cos^2 A + \sin^2 A = 1 \): \( = \frac{2 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = 2\sec A = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(iii). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \operatorname{cosec} \theta \)
Answer:
Write in terms of sine and cosine: \( \frac{\tan \theta}{1 - \cot \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} \) and \( \frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}} = \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)} \).
LHS \( = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)} = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta(\sin \theta - \cos \theta)} \).
Using \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \): \( = \frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = 1 + \sec \theta \operatorname{cosec} \theta = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(iv). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A} \)
Answer:
LHS \( = \frac{1 + \sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\cos A + 1}{1} = 1 + \cos A \).
RHS \( = \frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A \).
LHS = RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(v). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \operatorname{cosec} A + \cot A \), using the identity \( \operatorname{cosec}^2 A = 1 + \cot^2 A \)
Answer:
Divide the numerator and denominator of the LHS by \( \sin A \): LHS \( = \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A} \).
Replace 1 in the numerator by \( \operatorname{cosec}^2 A - \cot^2 A \): numerator \( = (\cot A + \operatorname{cosec} A) - (\operatorname{cosec}^2 A - \cot^2 A) = (\operatorname{cosec} A + \cot A) - (\operatorname{cosec} A + \cot A)(\operatorname{cosec} A - \cot A) \).
\( = (\operatorname{cosec} A + \cot A)(1 - \operatorname{cosec} A + \cot A) \). So LHS \( = \frac{(\operatorname{cosec} A + \cot A)(\cot A + 1 - \operatorname{cosec} A)}{\cot A + 1 - \operatorname{cosec} A} = \operatorname{cosec} A + \cot A = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(vi). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \)
Answer:
LHS \( = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \).
\( = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(vii). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta \)
Answer:
LHS \( = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)} \). Write \( 1 - 2\sin^2 \theta = \sin^2 \theta + \cos^2 \theta - 2\sin^2 \theta = \cos^2 \theta - \sin^2 \theta \) and \( 2\cos^2 \theta - 1 = 2\cos^2 \theta - \sin^2 \theta - \cos^2 \theta = \cos^2 \theta - \sin^2 \theta \).
LHS \( = \frac{\sin \theta(\cos^2 \theta - \sin^2 \theta)}{\cos \theta(\cos^2 \theta - \sin^2 \theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(viii). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( (\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)
Answer:
LHS \( = \sin^2 A + \operatorname{cosec}^2 A + 2\sin A \operatorname{cosec} A + \cos^2 A + \sec^2 A + 2\cos A \sec A \).
Since \( \sin A \operatorname{cosec} A = 1 \) and \( \cos A \sec A = 1 \): \( = (\sin^2 A + \cos^2 A) + 2 + 2 + \operatorname{cosec}^2 A + \sec^2 A = 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A) \).
\( = 7 + \tan^2 A + \cot^2 A = \) RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(ix). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( (\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A} \)
Answer:
LHS \( = \left( \frac{1}{\sin A} - \sin A \right)\left( \frac{1}{\cos A} - \cos A \right) = \frac{1 - \sin^2 A}{\sin A} \times \frac{1 - \cos^2 A}{\cos A} = \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A} = \sin A \cos A \).
RHS \( = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{\sin A \cos A}{1} = \sin A \cos A \).
LHS = RHS. Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.

Question 4(x). Prove the following identity, where the angles involved are acute angles for which the expressions are defined: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \left( \frac{1 - \tan A}{1 - \cot A} \right)^2 = \tan^2 A \)
Answer:
LHS \( = \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \).
Middle term: \( \frac{1 - \tan A}{1 - \cot A} = \frac{1 - \tan A}{1 - \frac{1}{\tan A}} = \frac{(1 - \tan A)\tan A}{\tan A - 1} = -\tan A \), so \( \left( \frac{1 - \tan A}{1 - \cot A} \right)^2 = \tan^2 A \).
All three expressions equal \( \tan^2 A \). Hence proved.
In simple words: Simplify one side with the basic identities until it matches the other side.

Teacher's Note:
1. Start with the more complicated side and change everything to sin and cos if you are stuck.
2. Stop as soon as the left side becomes exactly the right side.