CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 06

Prove the following identities:

Question. If \( \sec \theta = x + \frac{1}{4x} \), prove that \( \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \).
Answer: Since \( \tan \theta = \pm \sqrt{\sec^2 \theta - 1} \)
\( = \pm \sqrt{\left( x + \frac{1}{4x} \right)^2 - 1} \)
\( = \pm \left( x - \frac{1}{4x} \right) \)
Now, \( \sec \theta + \tan \theta = \left( x + \frac{1}{4x} \right) \pm \left( x - \frac{1}{4x} \right) \)
\( = 2x \) or \( \frac{1}{2x} \)

 

Question. If \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), show that \( m^2 - n^2 = 4 \sqrt{mn} \). 
Answer: LHS \( = m^2 - n^2 \)
\( = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 \) (Given)
\( = \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta - (\tan^2 \theta + \sin^2 \theta - 2 \tan \theta \sin \theta) \)
\( = 4 \tan \theta \sin \theta = 4 \sqrt{\tan^2 \theta \sin^2 \theta} = 4 \sqrt{\tan^2 \theta (1 - \cos^2 \theta)} \)
\( = 4 \sqrt{\tan^2 \theta - \sin^2 \theta} = 4 \sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)} \)
\( = 4 \sqrt{mn} \) (From given)
= RHS.

 

Question. If \( a \sec \theta + b \tan \theta + c = 0 \) and \( p \sec \theta + q \tan \theta + r = 0 \), prove that \( (br - qc)^2 - (pc - ar)^2 = (aq - pb)^2 \)
Answer: Given equation are
\( a \sec \theta + b \tan \theta + c = 0 \) ...(i)
\( p \sec \theta + q \tan \theta + r = 0 \) ...(ii)
Multiplying eq. (i) with \( p \) and eq. (ii) with \( a \), we get
\( ap \sec \theta + bp \tan \theta + cp = 0 \) ...(iii)
\( ap \sec \theta + aq \tan \theta + ar = 0 \) ...(iv)
Subtracting (iv) from (iii):
\( (bp - aq) \tan \theta + cp - ar = 0 \)

\( \implies \) \( \tan \theta = \frac{ar - cp}{bp - aq} = \frac{cp - ar}{aq - pb} \) ...(v)
Multiplying eq. (i) with \( q \) and eq. (ii) with \( b \), we get
\( aq \sec \theta + bq \tan \theta + cq = 0 \) ...(vi)
\( bp \sec \theta + bq \tan \theta + br = 0 \) ...(vii)
Subtracting (vii) from (vi):
\( (aq - bp) \sec \theta + cq - br = 0 \)

\( \implies \) \( \sec \theta = \frac{br - cq}{aq - pb} \)
Now \( \sec^2 \theta - \tan^2 \theta = 1 \)

\( \implies \) \( \left( \frac{br - cq}{aq - pb} \right)^2 - \left( \frac{cp - ar}{aq - pb} \right)^2 = 1 \)

\( \implies \) \( (br - cq)^2 - (cp - ar)^2 = (aq - pb)^2 \)

 

Long Answer Type Questions 

 

Question. Prove that: \( \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A \). 
Answer: LHS \( = \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} \)
\( = \frac{\sin A (1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)} \)
\( = \frac{\sin A (\sin^2 A + \cos^2 A - 2 \sin^2 A)}{\cos A (2 \cos^2 A - (\sin^2 A + \cos^2 A))} \)
\( = \frac{\sin A (\cos^2 A - \sin^2 A)}{\cos A (2 \cos^2 A - \sin^2 A - \cos^2 A)} \)
\( = \frac{\sin A (\cos^2 A - \sin^2 A)}{\cos A (\cos^2 A - \sin^2 A)} \)
\( = \frac{\sin A}{\cos A} = \tan A = \text{RHS} \). Hence proved.

 

Question. Prove that \( \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} = 2 \sec \theta \)
Answer: LHS \( = \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \)
\( = \frac{(\sqrt{1 + \sin \theta})^2 + (\sqrt{1 - \sin \theta})^2}{\sqrt{1 - \sin \theta} \times \sqrt{1 + \sin \theta}} = \frac{1 + \sin \theta + 1 - \sin \theta}{\sqrt{1 - \sin^2 \theta}} \)
\( = \frac{2}{\sqrt{\cos^2 \theta}} = \frac{2}{\cos \theta} = 2 \sec \theta = \text{RHS} \)

 

Question. If \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \) and \( x \sin \theta = y \cos \theta \), prove that \( x^2 + y^2 = 1 \). 
Answer: \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \) ...(i)
\( x \sin \theta - y \cos \theta = 0 \) ...(ii)
Equation (ii) \( \times \sin^2 \theta \) and subtracted from (i) we have
\( y (\cos^3 \theta + \cos \theta \sin^2 \theta) = \sin \theta \cos \theta \)

\( \implies \) \( y \cos \theta = \sin \theta \cos \theta \implies y = \sin \theta \)
Put in equation (ii), we get
\( x \sin \theta = \sin \theta \cos \theta \)

\( \implies \) \( x = \cos \theta \)

\( \implies \) \( x^2 + y^2 = \sin^2 \theta + \cos^2 \theta = 1 \)

 

Question. If \( a \cos \theta - b \sin \theta = c \), prove that \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \). 
Answer: \( a \cos \theta - b \sin \theta = c \)

\( \implies \) \( (a \cos \theta - b \sin \theta)^2 = c^2 \)

\( \implies \) \( a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \cos \theta \cdot \sin \theta = c^2 \)

\( \implies \) \( a^2(1 - \sin^2 \theta) + b^2(1 - \cos^2 \theta) - 2ab \cos \theta \cdot \sin \theta = c^2 \)

\( \implies \) \( a^2 - a^2 \sin^2 \theta + b^2 - b^2 \cos^2 \theta - 2ab \cos \theta \cdot \sin \theta = c^2 \)

\( \implies \) \( a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \cos \theta \cdot \sin \theta = a^2 + b^2 - c^2 \)

\( \implies \) \( (a \sin \theta + b \cos \theta)^2 = a^2 + b^2 - c^2 \)

\( \implies \) \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \)

 

Question. If \( \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \) and \( \frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta = 1 \), prove that \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \). 
Answer: \( \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1 \)
Squaring both sides, we get
\( \frac{x^2}{a^2} \cos^2 \theta + \frac{y^2}{b^2} \sin^2 \theta + \frac{2xy}{ab} \cos \theta \sin \theta = 1 \) ...(i)
and \( \frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta = 1 \)
Squaring both sides, we get
\( \frac{x^2}{a^2} \sin^2 \theta + \frac{y^2}{b^2} \cos^2 \theta - \frac{2xy}{ab} \cos \theta \sin \theta = 1 \) ...(ii)
Adding (i) and (ii), we get
\( \frac{x^2}{a^2} (\cos^2 \theta + \sin^2 \theta) + \frac{y^2}{b^2} (\sin^2 \theta + \cos^2 \theta) = 2 \)

\( \implies \) \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \) ( \(\because \sin^2 \theta + \cos^2 \theta = 1\) )
Hence proved.

 

Question. If \( x = a \cos^3 \theta \), \( y = b \sin^3 \theta \), prove that \( \left( \frac{x}{a} \right)^{2/3} + \left( \frac{y}{b} \right)^{2/3} = 1 \). 
Answer: \( x = a \cos^3 \theta, y = b \sin^3 \theta \)
LHS \( = \left( \frac{x}{a} \right)^{2/3} + \left( \frac{y}{b} \right)^{2/3} = \left( \frac{a \cos^3 \theta}{a} \right)^{2/3} + \left( \frac{b \sin^3 \theta}{b} \right)^{2/3} \)
\( = (\cos \theta)^{3 \times 2/3} + (\sin \theta)^{3 \times 2/3} = \cos^2 \theta + \sin^2 \theta = 1 \)
LHS = RHS. Hence proved.

 

Question. If \( \text{cosec } \theta - \sin \theta = m \) and \( \sec \theta - \cos \theta = n \), prove that \( (m^2n)^{2/3} + (mn^2)^{2/3} = 1 \). 
Answer: \( \text{cosec } \theta - \sin \theta = m \)

\( \implies \) \( \frac{1}{\sin \theta} - \sin \theta = m \implies \frac{1 - \sin^2 \theta}{\sin \theta} = m \)

\( \implies \) \( \frac{\cos^2 \theta}{\sin \theta} = m \) ...(i)
Also, \( \sec \theta - \cos \theta = n \)

\( \implies \) \( \frac{1}{\cos \theta} - \cos \theta = n \implies \frac{1 - \cos^2 \theta}{\cos \theta} = n \)

\( \implies \) \( \frac{\sin^2 \theta}{\cos \theta} = n \) ...(ii)
Now, LHS \( = (m^2n)^{2/3} + (mn^2)^{2/3} \)
\( = \left[ \left( \frac{\cos^2 \theta}{\sin \theta} \right)^2 \left( \frac{\sin^2 \theta}{\cos \theta} \right) \right]^{2/3} + \left[ \left( \frac{\cos^2 \theta}{\sin \theta} \right) \left( \frac{\sin^2 \theta}{\cos \theta} \right)^2 \right]^{2/3} \)
\( = \left( \frac{\cos^4 \theta \cdot \sin^2 \theta}{\sin^2 \theta \cdot \cos \theta} \right)^{2/3} + \left( \frac{\cos^2 \theta \cdot \sin^4 \theta}{\sin \theta \cdot \cos^2 \theta} \right)^{2/3} \)
\( = (\cos^3 \theta)^{2/3} + (\sin^3 \theta)^{2/3} = \cos^2 \theta + \sin^2 \theta = 1 = \text{RHS} \)

 

Practice Questions

 

Question. \( \sec A = \)
(a) \( \frac{1}{\cot A} \)
(b) \( \frac{1}{\text{cosec } A} \)
(c) \( \frac{1}{\sqrt{1 + \cot^2 A}} \)
(d) \( \frac{\sqrt{1 + \cot^2 A}}{\cot A} \)
Answer: (d) \( \frac{\sqrt{1 + \cot^2 A}}{\cot A} \)

 

Question. \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \)
(a) \( \tan^2 A \)
(b) \( \sec^2 A \)
(c) \( \text{cosec}^2 A - 1 \)
(d) \( 1 - \sin^2 A \)
Answer: (a) \( \tan^2 A \)

 

Question. If \( \sec A + \tan A = x \), then \( \tan A = \)
(a) \( \frac{x}{2} \)
(b) \( \frac{1}{2x} \)
(c) \( \frac{x^2 - 1}{2x} \)
(d) \( \frac{2x}{x^2 - 1} \)
Answer: (c) \( \frac{x^2 - 1}{2x} \)

 

Question. If \( \sin x + \sin^2 x = 1 \), then value of \( \cos^2 x + \cos^4 x = \)
(a) 1
(b) 2
(c) \( \frac{1}{2} \)
(d) 3
Answer: (a) 1

 

Question. If \( \sec^2 \theta (1 + \sin \theta) (1 - \sin \theta) = k \), then find the value of \( k \).
Answer: \( \sec^2 \theta (1 + \sin \theta) (1 - \sin \theta) = k \)
\( \implies \sec^2 \theta (1 - \sin^2 \theta) = k \)
\( \implies \sec^2 \theta \cdot \cos^2 \theta = k \)
\( \implies 1 = k \)
The value of \( k \) is 1.

 

Question. If \( 6x = \sec \theta \) and \( \frac{6}{x} = \tan \theta \), find the value of \( 9 \left( x^2 - \frac{1}{x^2} \right) \).
Answer: Given \( 6x = \sec \theta \implies x = \frac{\sec \theta}{6} \) and \( \frac{6}{x} = \tan \theta \implies \frac{1}{x} = \frac{\tan \theta}{6} \).
Now, \( 9 \left( x^2 - \frac{1}{x^2} \right) = 9 \left[ \left( \frac{\sec \theta}{6} \right)^2 - \left( \frac{\tan \theta}{6} \right)^2 \right] \)
\( \implies 9 \left[ \frac{\sec^2 \theta}{36} - \frac{\tan^2 \theta}{36} \right] \)
\( \implies \frac{9}{36} (\sec^2 \theta - \tan^2 \theta) \)
\( \implies \frac{1}{4} (1) = \frac{1}{4} \).
The value is \( \frac{1}{4} \).

 

Question. Simplify : \( \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \sin \theta \cos \theta \)
Answer: \( \frac{(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} + \sin \theta \cos \theta \)
\( \implies (\sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta) + \sin \theta \cos \theta \)
\( \implies (1 - \sin \theta \cos \theta) + \sin \theta \cos \theta = 1 \).
The simplified value is 1.

 

Question. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), show that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).
Answer: Given \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)
\( \implies \sin \theta = \sqrt{2} \cos \theta - \cos \theta \)
\( \implies \sin \theta = (\sqrt{2} - 1) \cos \theta \)
Multiplying both sides by \( (\sqrt{2} + 1) \):
\( (\sqrt{2} + 1) \sin \theta = (\sqrt{2} + 1)(\sqrt{2} - 1) \cos \theta \)
\( \implies \sqrt{2} \sin \theta + \sin \theta = (2 - 1) \cos \theta \)
\( \implies \sqrt{2} \sin \theta + \sin \theta = \cos \theta \)
\( \implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta \). Hence proved.

 

Question. Find the value of other trigonometric ratios, given that \( \tan \theta = \frac{2mn}{m^2 - n^2} \) 
Answer: Let perpendicular \( (p) = 2mn \) and base \( (b) = m^2 - n^2 \).
Then hypotenuse \( (h) = \sqrt{(2mn)^2 + (m^2 - n^2)^2} = \sqrt{4m^2n^2 + m^4 + n^4 - 2m^2n^2} = \sqrt{m^4 + n^4 + 2m^2n^2} = m^2 + n^2 \).
The other ratios are:
\( \sin \theta = \frac{2mn}{m^2 + n^2} \), \( \cos \theta = \frac{m^2 - n^2}{m^2 + n^2} \), \( \text{cosec } \theta = \frac{m^2 + n^2}{2mn} \), \( \sec \theta = \frac{m^2 + n^2}{m^2 - n^2} \), \( \cot \theta = \frac{m^2 - n^2}{2mn} \).

 

Question. Find the value of other trigonometric ratios, given that \( \sin \theta = \frac{m^2 - n^2}{m^2 + n^2} \) 
Answer: Let \( p = m^2 - n^2 \) and \( h = m^2 + n^2 \).
Then \( b = \sqrt{(m^2 + n^2)^2 - (m^2 - n^2)^2} = \sqrt{4m^2n^2} = 2mn \).
The other ratios are:
\( \cos \theta = \frac{2mn}{m^2 + n^2} \), \( \tan \theta = \frac{m^2 - n^2}{2mn} \), \( \text{cosec } \theta = \frac{m^2 + n^2}{m^2 - n^2} \), \( \sec \theta = \frac{m^2 + n^2}{2mn} \), \( \cot \theta = \frac{2mn}{m^2 - n^2} \).

 

Question. Find the value of other trigonometric ratios, given that \( \text{cosec } \theta = \sqrt{1 + \left( \frac{m}{n} \right)^4} \) 
Answer: \( \text{cosec } \theta = \sqrt{\frac{n^4 + m^4}{n^4}} = \frac{\sqrt{m^4 + n^4}}{n^2} \).
Let \( h = \sqrt{m^4 + n^4} \) and \( p = n^2 \).
Then \( b = \sqrt{(m^4 + n^4) - (n^2)^2} = \sqrt{m^4} = m^2 \).
Other ratios are:
\( \sin \theta = \frac{n^2}{\sqrt{m^4 + n^4}} \), \( \cos \theta = \frac{m^2}{\sqrt{m^4 + n^4}} \), \( \tan \theta = \frac{n^2}{m^2} \), \( \sec \theta = \frac{\sqrt{m^4 + n^4}}{m^2} \), \( \cot \theta = \frac{m^2}{n^2} \).

 

Question. Prove the following identity: \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{2\sin^2 A - 1} = \frac{2}{1 - 2\cos^2 A} \)
Answer: LHS \( = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{\sin^2 A - \cos^2 A} \)
\( \implies \frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A - (1 - \sin^2 A)} = \frac{2}{2\sin^2 A - 1} \).
Also \( \frac{2}{2(1 - \cos^2 A) - 1} = \frac{2}{2 - 2\cos^2 A - 1} = \frac{2}{1 - 2\cos^2 A} \). Hence proved.

 

Question. Prove the following identity: \( \sin^6 A + \cos^6 A = 1 - 3 \sin^2 A \cos^2 A \) 
Answer: LHS \( = (\sin^2 A)^3 + (\cos^2 A)^3 \)
\( \implies (\sin^2 A + \cos^2 A)(\sin^4 A - \sin^2 A \cos^2 A + \cos^4 A) \)
\( \implies (1)[(\sin^2 A + \cos^2 A)^2 - 2 \sin^2 A \cos^2 A - \sin^2 A \cos^2 A] \)
\( \implies (1)^2 - 3 \sin^2 A \cos^2 A = 1 - 3 \sin^2 A \cos^2 A = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \sin^8 \theta - \cos^8 \theta = (\sin^2 \theta - \cos^2 \theta) (1 - 2 \sin^2 \theta \cos^2 \theta) \)
Answer: LHS \( = (\sin^4 \theta)^2 - (\cos^4 \theta)^2 \)
\( \implies (\sin^4 \theta - \cos^4 \theta)(\sin^4 \theta + \cos^4 \theta) \)
\( \implies (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) [(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta] \)
\( \implies (\sin^2 \theta - \cos^2 \theta)(1) [1^2 - 2 \sin^2 \theta \cos^2 \theta] \)
\( \implies (\sin^2 \theta - \cos^2 \theta)(1 - 2 \sin^2 \theta \cos^2 \theta) = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \frac{(1 + \tan^2 \theta) \cot \theta}{\text{cosec}^2 \theta} = \tan \theta \)
Answer: LHS \( = \frac{\sec^2 \theta \cdot \cot \theta}{\text{cosec}^2 \theta} = \frac{\frac{1}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin^2 \theta}} \)
\( \implies \frac{1}{\cos \theta \sin \theta} \cdot \sin^2 \theta = \frac{\sin \theta}{\cos \theta} = \tan \theta = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \frac{1 + \cos A}{\sin A} = \frac{\sin A}{1 - \cos A} \)
Answer: LHS \( = \frac{1 + \cos A}{\sin A} \times \frac{1 - \cos A}{1 - \cos A} \)
\( \implies \frac{1 - \cos^2 A}{\sin A (1 - \cos A)} = \frac{\sin^2 A}{\sin A (1 - \cos A)} = \frac{\sin A}{1 - \cos A} = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \frac{1 + \cos \theta - \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta \)
Answer: LHS \( = \frac{(1 - \sin^2 \theta) + \cos \theta}{\sin \theta (1 + \cos \theta)} \)
\( \implies \frac{\cos^2 \theta + \cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{\cos \theta (\cos \theta + 1)}{\sin \theta (1 + \cos \theta)} \)
\( \implies \frac{\cos \theta}{\sin \theta} = \cot \theta = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \left( \tan \theta + \frac{1}{\cos \theta} \right)^2 + \left( \tan \theta - \frac{1}{\cos \theta} \right)^2 = 2 \left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) \)
Answer: LHS \( = (\tan \theta + \sec \theta)^2 + (\tan \theta - \sec \theta)^2 \)
\( \implies 2(\tan^2 \theta + \sec^2 \theta) = 2 \left( \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} \right) \)
\( \implies 2 \left( \frac{1 + \sin^2 \theta}{\cos^2 \theta} \right) = 2 \left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) = \) RHS. Hence proved.

 

Question. Prove the following identity: \( (\sec A + \tan A - 1) (\sec A - \tan A + 1) = 2 \tan A \) 
Answer: LHS \( = [\sec A + (\tan A - 1)][\sec A - (\tan A - 1)] \)
\( \implies \sec^2 A - (\tan A - 1)^2 = \sec^2 A - (\tan^2 A + 1 - 2 \tan A) \)
\( \implies \sec^2 A - \tan^2 A - 1 + 2 \tan A \)
\( \implies 1 - 1 + 2 \tan A = 2 \tan A = \) RHS. Hence proved.

 

Question. Prove the following identity: \( (\sec A - \text{cosec } A) (1 + \tan A + \cot A) = \tan A \sec A - \cot A \text{cosec } A \)
Answer: LHS \( = \left( \frac{1}{\cos A} - \frac{1}{\sin A} \right) \left( 1 + \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \right) \)
\( \implies \left( \frac{\sin A - \cos A}{\cos A \sin A} \right) \left( \frac{\sin A \cos A + \sin^2 A + \cos^2 A}{\sin A \cos A} \right) \)
\( \implies \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{\sin^2 A \cos^2 A} = \frac{\sin^3 A - \cos^3 A}{\sin^2 A \cos^2 A} \)
\( \implies \frac{\sin^3 A}{\sin^2 A \cos^2 A} - \frac{\cos^3 A}{\sin^2 A \cos^2 A} = \frac{\sin A}{\cos^2 A} - \frac{\cos A}{\sin^2 A} \)
\( \implies \tan A \sec A - \cot A \text{cosec } A = \) RHS. Hence proved.

 

Question. Prove the following identity: \( \sin^2 A \cos^2 B - \cos^2 A \sin^2 B = \sin^2 A - \sin^2 B \)
Answer: LHS \( = \sin^2 A (1 - \sin^2 B) - (1 - \sin^2 A) \sin^2 B \)
\( \implies \sin^2 A - \sin^2 A \sin^2 B - (\sin^2 B - \sin^2 A \sin^2 B) \)
\( \implies \sin^2 A - \sin^2 B = \) RHS. Hence proved.

 

Question. Prove the following identity: \( (\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \)
Answer: LHS \( = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \)
\( \implies \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \) RHS. Hence proved.

 

Question. If \( x = \gamma \cos \alpha \sin \beta \), \( y = \gamma \cos \alpha \cos \beta \) and \( z = \gamma \sin \alpha \), show that \( x^2 + y^2 + z^2 = \gamma^2 \). 
Answer: LHS \( = (\gamma \cos \alpha \sin \beta)^2 + (\gamma \cos \alpha \cos \beta)^2 + (\gamma \sin \alpha)^2 \)
\( \implies \gamma^2 \cos^2 \alpha (\sin^2 \beta + \cos^2 \beta) + \gamma^2 \sin^2 \alpha \)
\( \implies \gamma^2 \cos^2 \alpha (1) + \gamma^2 \sin^2 \alpha = \gamma^2 (\cos^2 \alpha + \sin^2 \alpha) = \gamma^2 = \) RHS. Hence proved.

 

Question. If \( \sin \theta - \cos \theta = \frac{1}{2} \), then find the value of \( \frac{1}{\sin \theta + \cos \theta} \). 
Answer: Squaring given: \( (\sin \theta - \cos \theta)^2 = 1/4 \implies 1 - 2\sin \theta \cos \theta = 1/4 \implies 2\sin \theta \cos \theta = 3/4 \).
Now, \( (\sin \theta + \cos \theta)^2 = 1 + 2\sin \theta \cos \theta = 1 + 3/4 = 7/4 \implies \sin \theta + \cos \theta = \frac{\sqrt{7}}{2} \).
Value of \( \frac{1}{\sin \theta + \cos \theta} = \frac{2}{\sqrt{7}} \).

 

Question. Solve the equation for \( \theta \): \( \frac{\cos^2 \theta}{\cot^2 \theta - \cos^2 \theta} = 3 \). 
Answer: \( \frac{\cos^2 \theta}{\frac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta} = 3 \implies \frac{\cos^2 \theta}{\cos^2 \theta \left( \frac{1 - \sin^2 \theta}{\sin^2 \theta} \right)} = 3 \)
\( \implies \frac{\sin^2 \theta}{\cos^2 \theta} = 3 \implies \tan^2 \theta = 3 \implies \tan \theta = \sqrt{3} \).
\( \implies \theta = 60^\circ \).

 

Question. Express \( \cos A \) in terms of \( \cot A \). 
Answer: \( \cos^2 A = 1 - \sin^2 A = 1 - \frac{1}{\text{cosec}^2 A} = 1 - \frac{1}{1 + \cot^2 A} = \frac{\cot^2 A}{1 + \cot^2 A} \).
\( \implies \cos A = \frac{\cot A}{\sqrt{1 + \cot^2 A}} \).

 

Question. Prove that : \( 2 (\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 = 0 \) 
Answer: LHS \( = 2(1 - 3 \sin^2 \theta \cos^2 \theta) - 3(1 - 2 \sin^2 \theta \cos^2 \theta) + 1 \)
\( \implies 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1 \)
\( \implies 2 - 3 + 1 = 0 = \) RHS. Hence proved.

 

Question. If \( 5 \sin \theta + 3 \cos \theta = 4 \), find the value of \( 3 \sin \theta - 5 \cos \theta \).
Answer: Let \( 3 \sin \theta - 5 \cos \theta = x \).
Squaring and adding both equations:
\( (5 \sin \theta + 3 \cos \theta)^2 + (3 \sin \theta - 5 \cos \theta)^2 = 4^2 + x^2 \)
\( \implies 25 \sin^2 \theta + 9 \cos^2 \theta + 30 \sin \theta \cos \theta + 9 \sin^2 \theta + 25 \cos^2 \theta - 30 \sin \theta \cos \theta = 16 + x^2 \)
\( \implies 34(\sin^2 \theta + \cos^2 \theta) = 16 + x^2 \implies 34 = 16 + x^2 \implies x^2 = 18 \).
\( \implies x = \pm 3\sqrt{2} \). Value is \( \pm 3\sqrt{2} \).

 

Question. Prove the following identity: \( \frac{1}{\cot^2 \theta} + \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 - \sin^2 \theta} - \frac{1}{\text{cosec}^2 \theta} \)
Answer: LHS \( = \tan^2 \theta + \cos^2 \theta = \sec^2 \theta - 1 + \cos^2 \theta \).
RHS \( = \frac{1}{\cos^2 \theta} - \sin^2 \theta = \sec^2 \theta - (1 - \cos^2 \theta) = \sec^2 \theta - 1 + \cos^2 \theta \).
LHS = RHS. Hence proved.

 

Question. Prove the following identity: \( \frac{\sin A}{\sec A + \tan A - 1} + \frac{\cos A}{\text{cosec } A + \cot A - 1} = 1 \)
Answer: LHS \( = \frac{\sin A}{\frac{1 + \sin A - \cos A}{\cos A}} + \frac{\cos A}{\frac{1 + \cos A - \sin A}{\sin A}} = \frac{\sin A \cos A}{1 + \sin A - \cos A} + \frac{\sin A \cos A}{1 + \cos A - \sin A} \)
\( \implies \sin A \cos A \left[ \frac{(1 + \cos A - \sin A) + (1 + \sin A - \cos A)}{(1 + \sin A - \cos A)(1 - (\sin A - \cos A))} \right] \)
\( \implies \sin A \cos A \left[ \frac{2}{1 - (\sin A - \cos A)^2} \right] = \sin A \cos A \left[ \frac{2}{1 - (1 - 2 \sin A \cos A)} \right] \)
\( \implies \sin A \cos A \cdot \frac{2}{2 \sin A \cos A} = 1 = \) RHS. Hence proved.

 

Question. If \( \text{cosec } \theta - \sin \theta = l \) and \( \sec \theta - \cos \theta = m \), show that \( l^2 m^2 (l^2 + m^2 + 3) = 1 \).
Answer: \( l = \frac{\cos^2 \theta}{\sin \theta} \) and \( m = \frac{\sin^2 \theta}{\cos \theta} \).
\( l^2 m^2 = \cos^2 \theta \sin^2 \theta \).
\( l^2 + m^2 + 3 = \frac{\cos^4 \theta}{\sin^2 \theta} + \frac{\sin^4 \theta}{\cos^2 \theta} + 3 = \frac{\cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} \).
LHS \( = (\cos^2 \theta \sin^2 \theta) \left( \frac{1}{\sin^2 \theta \cos^2 \theta} \right) = 1 = \) RHS. Hence proved.

 

Question. If \( \cos A - \sin A = m \) and \( \cos A + \sin A = n \). Show that: \( \frac{m^2 - n^2}{m^2 + n^2} = -2 \sin A \cos A = - \frac{2}{\tan A + \cot A} \). 
Answer: \( m^2 - n^2 = (m - n)(m + n) = (-2 \sin A)(2 \cos A) = -4 \sin A \cos A \).
\( m^2 + n^2 = 2(\cos^2 A + \sin^2 A) = 2 \).
\( \frac{m^2 - n^2}{m^2 + n^2} = \frac{-4 \sin A \cos A}{2} = -2 \sin A \cos A \).
Also, \( -2 \sin A \cos A = \frac{-2}{\frac{1}{\sin A \cos A}} = \frac{-2}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{-2}{\tan A + \cot A} \). Hence proved.

 

Question. Prove that: \( \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1} = 2 \text{cosec } A \). 
Answer: LHS \( = \tan A \left[ \frac{\sec A + 1 + \sec A - 1}{\sec^2 A - 1} \right] = \tan A \cdot \frac{2 \sec A}{\tan^2 A} \)
\( \implies \frac{2 \sec A}{\tan A} = \frac{2/\cos A}{\sin A / \cos A} = \frac{2}{\sin A} = 2 \text{cosec } A = \) RHS. Hence proved.

 

Question. If \( \text{cosec } A + \cot A = m \). Show that \( \frac{m^2 - 1}{m^2 + 1} = \cos A \). 
Answer: \( m = \frac{1 + \cos A}{\sin A} \implies m^2 = \frac{(1 + \cos A)^2}{1 - \cos^2 A} = \frac{1 + \cos A}{1 - \cos A} \).
LHS \( = \frac{\frac{1 + \cos A}{1 - \cos A} - 1}{\frac{1 + \cos A}{1 - \cos A} + 1} = \frac{1 + \cos A - 1 + \cos A}{1 + \cos A + 1 - \cos A} = \frac{2 \cos A}{2} = \cos A = \) RHS. Hence proved.

 

Question. If \( \sec \theta + \tan \theta = p \), find the value of \( \text{cosec } \theta \). 
Answer: Given \( \sec \theta + \tan \theta = p \). Then \( \sec \theta - \tan \theta = \frac{1}{p} \).
Adding: \( 2 \sec \theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \cos \theta = \frac{2p}{p^2 + 1} \).
Subtracting: \( 2 \tan \theta = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \sin \theta = \frac{p^2 - 1}{p^2 + 1} \) (since \( \sin \theta = \tan \theta \cdot \cos \theta \)).
Value of \( \text{cosec } \theta = \frac{p^2 + 1}{p^2 - 1} \).

 

 

Trignometry

Q.- Prove that 

(1 – sinθ + cosθ)² = 2(1 + cosθ)(1 – sinθ)

Sol. (1 – sinθ + cosθ)²

= 1 + sin²θ + cos²θ – 2sinθ + 2cosθ– 2sinθcosθ

= 2 – 2sinθ + 2cosθ – 2sinθ cosθ

= 2 (1 – sinθ) + 2 cosθ (1 – sinθ) 

 

 

 

Please click on below link to download CBSE Class 10 Mathematics Trignometry Printable Worksheet Set F