CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 05

Question. If \(0^\circ < \theta < 90^\circ\), then \(\sec \theta\) is
(a) \(> 1\)
(b) \(< 1\)
(c) \(= 1\)
(d) 0
Answer: (a) > 1

 

Question. If \(\sin \theta = \sqrt{3} \cos \theta\), \(0^\circ < \theta < 90^\circ\), then \(\theta\) is equal to
(a) \(30^\circ\)
(b) \(45^\circ\)
(c) \(60^\circ\)
(d) \(90^\circ\)
Answer: (c) 60°

 

Question. If \(\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3\), \(0^\circ < \theta_1, \theta_2, \theta_3 \le 90^\circ\), then find \(\cos \theta_1 + \cos \theta_2 + \cos \theta_3\).
Answer: Since the maximum value of \(\sin \theta\) is 1 and their sum is 3, then \(\sin \theta_1 = \sin \theta_2 = \sin \theta_3 = 1\). This implies \(\theta_1 = \theta_2 = \theta_3 = 90^\circ\).
\(\therefore \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = \cos 90^\circ + \cos 90^\circ + \cos 90^\circ = 0 + 0 + 0 = 0\).

 

Question. If \(\tan \frac{5\theta}{2} = \sqrt{3}\) and \(\theta\) is acute, then find the value of \(2\theta\). 
Answer: \(\tan \frac{5\theta}{2} = \tan 60^\circ \implies \frac{5\theta}{2} = 60^\circ \implies 5\theta = 120^\circ \implies \theta = 24^\circ\).
Value of \(2\theta = 2 \times 24^\circ = 48^\circ\).

 

Question. If \(\sqrt{3} \sin \theta = \cos \theta\), find the value of \(\frac{3\cos^2 \theta + 2\cos \theta}{3\cos \theta + 2}\). 
Answer: \(\sqrt{3} \sin \theta = \cos \theta \implies \tan \theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ\).
Expression \( = \frac{\cos \theta(3\cos \theta + 2)}{3\cos \theta + 2} = \cos \theta\).
For \(\theta = 30^\circ\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\).

 

Question. If \(\sin A = \frac{\sqrt{3}}{2}\), find the value of \(2\cot^2 A - 1\). 
Answer: \(\sin A = \frac{\sqrt{3}}{2} \implies A = 60^\circ\).
Value \( = 2\cot^2 60^\circ - 1 = 2 \times \left(\frac{1}{\sqrt{3}}\right)^2 - 1 = 2 \times \frac{1}{3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3}\).

 

Question. Find the value of \(\theta\) (\(0^\circ < \theta < 90^\circ\)) if \(2\cos^2 \theta = \frac{1}{2}\).
Answer: \(2\cos^2 \theta = \frac{1}{2} \implies \cos^2 \theta = \frac{1}{4} \implies \cos \theta = \frac{1}{2}\).
\( \implies \) \(\theta = 60^\circ\).

 

Question. Find the value of \(\theta\) if \(\sqrt{3} \tan 2\theta - 3 = 0\).
Answer: \(\sqrt{3} \tan 2\theta = 3 \implies \tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3}\).
\( \implies \) \(2\theta = 60^\circ \implies \theta = 30^\circ\).

 

Question. \(ABC\) is a triangle right angled at \(C\) and \(AC = \sqrt{3} BC\), prove that \(\angle ABC = 60^\circ\). 
Answer: In \(\Delta ABC\), \(\tan B = \frac{AC}{BC} = \frac{\sqrt{3}BC}{BC} = \sqrt{3}\).
\( \implies \) \(\angle ABC = 60^\circ\).

 

Question. If \(\sin A = \cos A\), find the value of \(2\tan^2 A + \sin^2 A - 1\). 
Answer: \(\sin A = \cos A \implies \tan A = 1 \implies A = 45^\circ\).
Value \( = 2\tan^2 45^\circ + \sin^2 45^\circ - 1 = 2(1)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 - 1 = 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{3}{2}\).

 

Question. Determine the value of \(x\) such that \(2 \text{cosec}^2 30^\circ + x \sin^2 60^\circ - \frac{3}{4} \tan^2 30^\circ = 10\)
Answer: \(2(2)^2 + x\left(\frac{\sqrt{3}}{2}\right)^2 - \frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2 = 10\)
\( \implies \) \(2(4) + x\left(\frac{3}{4}\right) - \frac{3}{4}\left(\frac{1}{3}\right) = 10\)
\( \implies \) \(8 + \frac{3x}{4} - \frac{1}{4} = 10\)
\( \implies \) \(\frac{3x - 1}{4} = 2\)
\( \implies \) \(3x - 1 = 8 \implies 3x = 9 \implies x = 3\).

 

Question. In an acute angled triangle \(ABC\), if \(\sin (A + B - C) = \frac{1}{2}\) and \(\cos (B + C - A) = \frac{1}{\sqrt{2}}\), find \(\angle A\), \(\angle B\) and \(\angle C\). 
Answer: \(A + B - C = 30^\circ\) ...(i)
\(B + C - A = 45^\circ\) ...(ii)
Also, \(A + B + C = 180^\circ\) ...(iii)
Adding (i) and (ii): \(2B = 75^\circ \implies B = 37.5^\circ\).
Subtracting (ii) from (iii): \(2A = 135^\circ \implies A = 67.5^\circ\).
From (iii), \(C = 180^\circ - (67.5^\circ + 37.5^\circ) = 75^\circ\).

 

Question. Given that \(\cos (A - B) = \cos A \cos B + \sin A \sin B\), find the value of \(\cos 15^\circ\) in two ways.
(i) Taking \(A = 60^\circ\), \(B = 45^\circ\) and
(ii) taking \(A = 45^\circ\) and \(B = 30^\circ\) 
Answer: (i) \(\cos(60^\circ - 45^\circ) = \cos 60^\circ \cos 45^\circ + \sin 60^\circ \sin 45^\circ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{3}}{2\sqrt{2}}\).
(ii) \(\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}\).

 

Question. If \(\theta = 30^\circ\), verify the following:
(i) \(\cos 3\theta = 4\cos^3 \theta - 3\cos \theta\)
(ii) \(\sin 3\theta = 3\sin \theta - 4\sin^3 \theta\) 
Answer: (i) LHS \(= \cos 90^\circ = 0\). RHS \(= 4\cos^3 30^\circ - 3\cos 30^\circ = 4(\frac{\sqrt{3}}{2})^3 - 3(\frac{\sqrt{3}}{2}) = 4 \cdot \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = 0\).
(ii) LHS \(= \sin 90^\circ = 1\). RHS \(= 3\sin 30^\circ - 4\sin^3 30^\circ = 3(\frac{1}{2}) - 4(\frac{1}{2})^3 = \frac{3}{2} - \frac{4}{8} = 1\).

 

Very Short Answer Type Questions 

 

Question. If \(7 \sin^2 \theta + 3 \cos^2 \theta = 4\), then show that \(\tan \theta = \frac{1}{\sqrt{3}}\).
Answer: \(7 \sin^2 \theta + 3 \cos^2 \theta = 4 \implies 4 \sin^2 \theta + 3(\sin^2 \theta + \cos^2 \theta) = 4\)
\( \implies \) \(4 \sin^2 \theta + 3 = 4 \implies 4 \sin^2 \theta = 1\)
\( \implies \) \(\sin^2 \theta = \frac{1}{4} \implies \sin \theta = \frac{1}{2}\)
\( \implies \) \(\theta = 30^\circ\)
\(\therefore \tan 30^\circ = \frac{1}{\sqrt{3}}\)

 

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

 

Question. If \(\sec \theta + \tan \theta = m\) and \(\sec \theta - \tan \theta = n\), find the value of \(\sqrt{mn}\).
Answer: Multiplying the given equations:
\((\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = mn\)
\( \implies \) \(\sec^2 \theta - \tan^2 \theta = mn\)
\( \implies \) \(1 = mn \implies \sqrt{mn} = 1\).

 

Question. If \(\sin \theta + \cos \theta = p\) and \(\sec \theta + \text{cosec } \theta = q\), show that \(q(p^2 - 1) = 2p\). 
Answer: LHS \( = q(p^2 - 1) = (\sec \theta + \text{cosec } \theta)[(\sin \theta + \cos \theta)^2 - 1]\)
\( = \left(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right)[\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1]\)
\( = \left(\frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta}\right)[1 + 2\sin \theta \cos \theta - 1]\)
\( = \frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta} \times 2\sin \theta \cos \theta\)
\( = 2(\sin \theta + \cos \theta) = 2p = \) RHS. Hence proved.

 

Question. Show that : \(\left(1 + \frac{1}{\tan^2 \theta}\right)\left(1 + \frac{1}{\cot^2 \theta}\right) = \frac{1}{\sin^2 \theta - \sin^4 \theta}\)
Answer: LHS \( = (1 + \cot^2 \theta)(1 + \tan^2 \theta)\)
\( = \text{cosec}^2 \theta \cdot \sec^2 \theta\)
\( = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta(1 - \sin^2 \theta)}\)
\( = \frac{1}{\sin^2 \theta - \sin^4 \theta} = \) RHS.

Question. If \( \cot \theta = \sqrt{7} \), show that \( \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} = \frac{3}{4} \).
Answer: LHS \( = \frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} \)
Dividing the numerator and denominator by \( \sec^2 \theta \). We get
\( \frac{\cot^2 \theta - 1}{\cot^2 \theta + 1} = \frac{(\sqrt{7})^2 - 1}{(\sqrt{7})^2 + 1} = \frac{7 - 1}{7 + 1} = \frac{6}{8} = \frac{3}{4} = \text{RHS} \)

 

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

 

Question. Prove the following identity: \( \frac{\tan \theta - \cot \theta}{\sin \theta \cdot \cos \theta} = \tan^2 \theta - \cot^2 \theta \)
Answer: RHS \( = \tan^2 \theta - \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta} \)
\( = \frac{\sin^4 \theta - \cos^4 \theta}{\sin^2 \theta \cdot \cos^2 \theta} = \frac{(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)}{\sin^2 \theta \cdot \cos^2 \theta} \)
\( = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cdot \cos^2 \theta} \left[ \because \sin^2 \theta + \cos^2 \theta = 1 \right] \)
\( = \frac{\tan \theta - \cot \theta}{\sin \theta \cdot \cos \theta} = \text{LHS} \).

 

Question. Prove the following identity: \( \frac{\sec \theta + \tan \theta}{\sec \theta - \tan \theta} = 1 - 2 \sec \theta \cdot \tan \theta + 2 \tan^2 \theta \)
Answer: LHS \( = \frac{\sec \theta + \tan \theta}{\sec \theta - \tan \theta} = \frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} \times \frac{\sec \theta - \tan \theta}{\sec \theta - \tan \theta} \)
\( = \frac{(\sec \theta - \tan \theta)^2}{\sec^2 \theta - \tan^2 \theta} = \frac{\sec^2 \theta + \tan^2 \theta - 2 \sec \theta \cdot \tan \theta}{1} \)
\( = 1 + \tan^2 \theta + \tan^2 \theta - 2 \sec \theta \cdot \tan \theta \)
\( = 1 + 2 \tan^2 \theta - 2 \sec \theta \cdot \tan \theta = \text{RHS} \).

 

Question. Prove the following identity : \( \cos^4 A - \cos^2 A = \sin^4 A - \sin^2 A \) 
Answer: LHS \( = \cos^4 A - \cos^2 A \)
\( = \cos^2 A (\cos^2 A - 1) \)
\( = (1 - \sin^2 A) (-\sin^2 A) \)
\( = \sin^4 A - \sin^2 A = \text{RHS} \).

 

Question. Prove the following identity: \( \tan^2 \theta - \sin^2 \theta = \tan^2 \theta \cdot \sin^2 \theta \)
Answer: LHS \( = \tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta \)
\( = \frac{\sin^2 \theta - \cos^2 \theta \cdot \sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta} \)
\( = \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) \cdot \sin^2 \theta = \tan^2 \theta \cdot \sin^2 \theta = \text{RHS} \)

 

Question. Prove the following identity: \( \sec^6 \theta = \tan^6 \theta + 3 \tan^2 \theta \sec^2 \theta + 1 \)
Answer: LHS \( = \sec^6 \theta = (\sec^2 \theta)^3 = (1 + \tan^2 \theta)^3 \)
\( = (1)^3 + (\tan^2 \theta)^3 + 3 \cdot 1 \cdot \tan^2 \theta(1 + \tan^2 \theta) \)
\( = 1 + \tan^6 \theta + 3\tan^2 \theta \cdot \sec^2 \theta = \text{RHS} \)

 

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

 

Question. Prove the following identity : \( \frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \cdot \tan B \)
Answer: LHS \( = \frac{\tan A + \tan B}{\cot A + \cot B} = \frac{\tan A + \tan B}{\frac{1}{\tan A} + \frac{1}{\tan B}} \)
\( = \frac{\tan A + \tan B}{\frac{\tan B + \tan A}{\tan A \cdot \tan B}} = \frac{(\tan A + \tan B) \times \tan A \cdot \tan B}{(\tan A + \tan B)} \)
\( = \tan A \cdot \tan B = \text{RHS} \)

 

Question. Prove the following identity : If \( \cos \theta - \sin \theta = 1 \), show that \( \cos \theta + \sin \theta = 1 \) or \( - 1 \).
Answer: \( (\cos \theta - \sin \theta)^2 = (1)^2 \)
\( \implies \) \( \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cdot \cos \theta = 1 \)
\( \implies \) \( 2 \sin \theta \cdot \cos \theta = 0 \) ...(i)
Now, \( (\cos \theta + \sin \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cdot \cos \theta \)
\( \implies \) \( (\cos \theta + \sin \theta)^2 = 1 + 0 \) [Using (i)]
\( \implies \) \( \cos \theta + \sin \theta = \pm \sqrt{1} = \pm 1 \)

 

Question. Prove the following identity : If \( x = a \sec \theta \), \( y = b \tan \theta \), prove that \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Answer: \( x = a \sec \theta \implies \frac{x}{a} = \sec \theta \)
\( \implies \) \( \frac{x^2}{a^2} = \sec^2 \theta \) ...(i)
Also \( y = b \tan \theta \implies \frac{y}{b} = \tan \theta \)
\( \implies \) \( \frac{y^2}{b^2} = \tan^2 \theta \) ...(ii)
Subtracting equation (ii) from (i), we get
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = \sec^2 \theta - \tan^2 \theta \)
\( \implies \) \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) Hence proved.

 

Question. If \( a \cos \theta - b \sin \theta = x \) and \( a \sin \theta + b \cos \theta = y \). Prove that \( a^2 + b^2 = x^2 + y^2 \).
Answer: Given. \( a \cos \theta - b \sin \theta = x \) and \( a \sin \theta + b \cos \theta = y \)
To show. \( a^2 + b^2 = x^2 + y^2 \)
Sol. RHS \( = x^2 + y^2 \)
\( = (a \cos \theta - b \sin \theta)^2 + (a \sin \theta + b \cos \theta)^2 \)
\( = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \cos \theta \sin \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta \)
\( = a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\cos^2 \theta + \sin^2 \theta) \)
\( = a^2 (1) + b^2(1) = a^2 + b^2 = \text{LHS} \)

 

Short Answer Type Questions 

 

Question. Prove that: \( \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} + \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{2\sec^2 \theta}{\tan^2 \theta - 1} \)
Answer: LHS \( = \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} + \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} \)
\( = \frac{(\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2}{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta + \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta}{\sin^2 \theta - \cos^2 \theta} \)
\( = \frac{2}{\sin^2 \theta - \cos^2 \theta} = \frac{\frac{2}{\cos^2 \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\cos^2 \theta}{\cos^2 \theta}} \) [Dividing Numerator and Denominator by \( \cos^2 \theta \)]
\( = \frac{2 \sec^2 \theta}{\tan^2 \theta - 1} = \text{RHS} \)

 

Question. Prove that: \( \frac{1}{\text{cosec } \theta - \cot \theta} - \frac{1}{\sin \theta} = \frac{1}{\sin \theta} - \frac{1}{\text{cosec } \theta + \cot \theta} \)
Answer: LHS \( = \frac{1}{\text{cosec } \theta - \cot \theta} - \frac{1}{\sin \theta} \)
\( = \frac{\text{cosec } \theta + \cot \theta}{(\text{cosec } \theta - \cot \theta)(\text{cosec } \theta + \cot \theta)} - \text{cosec } \theta \)
\( = \frac{\text{cosec } \theta + \cot \theta}{\text{cosec}^2 \theta - \cot^2 \theta} - \text{cosec } \theta \)
\( = \text{cosec } \theta + \cot \theta - \text{cosec } \theta = \text{cosec } \theta - (\text{cosec } \theta - \cot \theta) \)
\( = \frac{1}{\sin \theta} - \frac{(\text{cosec } \theta - \cot \theta)(\text{cosec } \theta + \cot \theta)}{\text{cosec } \theta + \cot \theta} \)
\( = \frac{1}{\sin \theta} - \frac{\text{cosec}^2 \theta - \cot^2 \theta}{\text{cosec } \theta + \cot \theta} = \frac{1}{\sin \theta} - \frac{1}{\text{cosec } \theta + \cot \theta} = \text{RHS} \)

 

Question. Prove that: \( \sin \theta (1 + \tan \theta) + \cos \theta (1 + \cot \theta) = \sec \theta + \text{cosec } \theta \).
Answer: LHS \( = \sin \theta (1 + \tan \theta) + \cos \theta (1 + \cot \theta) \)
\( = \sin \theta + \sin \theta \cdot \tan \theta + \cos \theta + \cos \theta \cdot \cot \theta \)
\( = \sin \theta + \sin \theta \cdot \frac{\sin \theta}{\cos \theta} + \cos \theta + \cos \theta \cdot \frac{\cos \theta}{\sin \theta} \)
\( = \sin \theta + \frac{\sin^2 \theta}{\cos \theta} + \cos \theta + \frac{\cos^2 \theta}{\sin \theta} \)
\( = \frac{\sin \theta \cos \theta + \sin^2 \theta}{\cos \theta} + \frac{\sin \theta \cos \theta + \cos^2 \theta}{\sin \theta} \)
\( = \frac{1}{\sin \theta} + \frac{1}{\cos \theta} \)
\( = \text{cosec } \theta + \sec \theta = \sec \theta + \text{cosec } \theta = \text{RHS} \)

 

Question. Prove that : \( \sec^2 \theta - \left[ \frac{\sin^2 \theta - 2\sin^4 \theta}{2\cos^4 \theta - \cos^2 \theta} \right] = 1 \)
Answer: LHS \( = \sec^2 \theta - \frac{\sin^2 \theta - 2\sin^4 \theta}{2\cos^4 \theta - \cos^2 \theta} \)
\( = \sec^2 \theta - \frac{\sin^2 \theta (1 - 2\sin^2 \theta)}{\cos^2 \theta (2\cos^2 \theta - 1)} \)
\( = \sec^2 \theta - \frac{\sin^2 \theta (\cos^2 \theta - \sin^2 \theta)}{\cos^2 \theta (\cos^2 \theta - \sin^2 \theta)} \)
\( = \sec^2 \theta - \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta} \)
\( = \frac{1 - \sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta}{\cos^2 \theta} = 1 = \text{RHS} \) Hence proved.

 

Question. \( (\tan A + \text{cosec } B)^2 - (\cot B - \sec A)^2 = 2 \tan A \cot B (\text{cosec } A + \sec B) \)
Answer: LHS \( = (\tan A + \text{cosec } B)^2 - (\cot B - \sec A)^2 \)
\( = \tan^2 A + \text{cosec}^2 B + 2 \tan A \cdot \text{cosec } B - (\cot^2 B + \sec^2 A - 2 \cot B \cdot \sec A) \)
\( = \tan^2 A + \text{cosec}^2 B + 2 \tan A \cdot \text{cosec } B - \cot^2 B - \sec^2 A + 2 \cot B \sec A \)
\( = (\text{cosec}^2 B - \cot^2 B) - (\sec^2 A - \tan^2 A) + 2 \tan A \cdot \text{cosec } B + 2 \cot B \sec A \)
\( = 1 - 1 + 2 \tan A \cdot \text{cosec } B + 2 \cot B \cdot \sec A \)
\( = 2 \tan A \cdot \text{cosec } B + 2 \cot B \cdot \sec A \)
\( = 2 \left( \frac{\sin A}{\cos A} \times \frac{1}{\sin B} + \frac{\cos B}{\sin B} \times \frac{1}{\cos A} \right) \)
\( = 2 \frac{\sin A + \cos B}{\cos A \cdot \sin B} \) ...(i)
RHS \( = 2 \tan A \cdot \cot B \cdot (\text{cosec } A + \sec B) \)
\( = 2 \tan A \cdot \cot B \text{cosec } A + 2 \tan A \cot B \cdot \sec B \)
\( = 2 \frac{\sin A}{\cos A} \cdot \frac{\cos B}{\sin B} \cdot \frac{1}{\sin A} + 2 \frac{\sin A}{\cos A} \cdot \frac{\cos B}{\sin B} \cdot \frac{1}{\cos B} \)
\( = 2 \left[ \frac{\cos B}{\cos A \cdot \sin B} + \frac{\sin A}{\cos A \cdot \sin B} \right] \)
\( = 2 \left[ \frac{\cos B + \sin A}{\cos A \cdot \sin B} \right] \) ...(ii)
From (i) and (ii),
LHS = RHS
Hence proved.

 

Question. \( (\sin A + \sec A)^2 + (\cos A + \text{cosec } A)^2 = (1 + \sec A \cdot \text{cosec } A)^2 \)
Answer: LHS \( = (\sin A + \sec A)^2 + (\cos A + \text{cosec } A)^2 \)
\( = \left( \sin A + \frac{1}{\cos A} \right)^2 + \left( \cos A + \frac{1}{\sin A} \right)^2 \)
\( = \left( \frac{\sin A \cos A + 1}{\cos A} \right)^2 + \left( \frac{\cos A \sin A + 1}{\sin A} \right)^2 \)
\( = \frac{\sin^2 A (\sin A \cos A + 1)^2 + \cos^2 A (\cos A \sin A + 1)^2}{\cos^2 A \sin^2 A} \)
\( = \frac{(\sin A \cos A + 1)^2 (\sin^2 A + \cos^2 A)}{\cos^2 A \sin^2 A} \)
\( = \frac{\sin^2 A \cos^2 A + 1 + 2 \sin A \cos A}{\cos^2 A \sin^2 A} \)
\( = \frac{\sin^2 A \cos^2 A}{\cos^2 A \sin^2 A} + \frac{1}{\cos^2 A \sin^2 A} + \frac{2 \sin A \cos A}{\cos^2 A \sin^2 A} \)
\( = 1 + \sec^2 A \text{cosec}^2 A + 2 \sec A \text{cosec } A \)
\( = (1 + \sec A \text{cosec } A)^2 = \text{RHS} \)

 

Question. \( \cot^2 A \left( \frac{\sec A - 1}{1 + \sin A} \right) + \sec^2 A \left( \frac{\sin A - 1}{1 + \sec A} \right) = 0 \)
Answer: Simplify each term separately
\( \frac{\cos^2 A}{\sin^2 A} \left( \frac{\sec A - 1}{1 + \sin A} \right) + \sec^2 A \left( \frac{\sin A - 1}{1 + \sec A} \right) \)
\( = \frac{(1 - \sin A)(\sec A - 1)}{(1 - \cos^2 A)} + \sec^2 A \left( \frac{\sin A - 1}{1 + \sec A} \right) \)
\( = \frac{(1 - \sin A) \sec^2 A}{(1 + \sec A)} - \frac{\sec^2 A (1 - \sin A)}{1 + \sec A} = 0 \)

 

Question. \( \frac{1}{\sec A - 1} + \frac{1}{\sec A + 1} = 2 \text{cosec } A \cdot \cot A \)
Answer: LHS \( = \frac{1}{\sec A - 1} + \frac{1}{\sec A + 1} \)
\( = \frac{\sec A + 1 + \sec A - 1}{(\sec A - 1)(\sec A + 1)} = \frac{2 \sec A}{\sec^2 A - 1} \)
\( = \frac{2 \sec A}{\tan^2 A} = \frac{2 \times \frac{1}{\cos A}}{\frac{\sin^2 A}{\cos^2 A}} = \frac{2 \times 1}{\cos A} \times \frac{\cos^2 A}{\sin^2 A} \)
\( = 2 \times \cos A \times \frac{1}{\sin A \cdot \sin A} \)
\( = 2 \cot A \cdot \text{cosec } A = \text{RHS} \)

 

Question. Show that \( 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) \) is independent of \( \theta \). 
Answer: \( 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) \)
\( = 3[(\sin \theta - \cos \theta)^2]^2 + 6(\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cdot \cos \theta) + 4[(\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cdot \cos^2 \theta(\sin^2 \theta + \cos^2 \theta)] \)
\( = 3[\sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cdot \cos \theta]^2 + 6[1 + 2 \sin \theta \cdot \cos \theta] + 4(1 - 3 \sin^2 \theta \cdot \cos^2 \theta) \)
\( = 3(1 + 4 \sin^2 \theta \cdot \cos^2 \theta - 4 \sin \theta \cos \theta) + 6 + 12 \sin \theta \cdot \cos \theta + 4 - 12 \sin^2 \theta \cos^2 \theta \)
\( = 3 + 12 \sin^2 \theta \cos^2 \theta - 12 \sin \theta \cos \theta + 6 + 12 \sin \theta \cos \theta + 4 - 12 \sin^2 \theta \cos^2 \theta \)
\( = 13 \) \( \implies \) independent of \( \theta \).

Trignometry 

 

 

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