CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 02

Question. If \( \tan (A + B) = \sqrt{3} \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \); \( 0^\circ < A + B \le 90^\circ; A > B \), find \( A \) and \( B \).
Answer: We have, \( \tan (A + B) = \sqrt{3} \)
\( \implies \) \( \tan (A + B) = \tan 60^\circ \)
\( \therefore A + B = 60^\circ \) ... (i)
Again, \( \tan (A - B) = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \tan (A - B) = \tan 30^\circ \)
\( \therefore A - B = 30^\circ \) ... (ii)
Adding (i) and (ii), we have
\( 2A = 90^\circ \)
\( \implies \) \( A = 45^\circ \)
Putting the value of \( A \) in (i), we have
\( 45^\circ + B = 60^\circ \)
\( \therefore B = 60^\circ - 45^\circ = 15^\circ \)
Hence, \( A = 45^\circ \) and \( B = 15^\circ \).

Question. Prove that: \( (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \) 
Answer: LHS \( = (\sin A + \csc A)^2 + (\cos A + \sec A)^2 \)
\( = \sin^2 A + \csc^2 A + 2\sin A . \csc A + \cos^2 A + \sec^2 A + 2 \cos A . \sec A \)
\( = (\sin^2 A + \csc^2 A + 2) + (\cos^2 A + \sec^2 A + 2) \) (\( \because \sin A . \csc A = 1 \) & \( \cos A . \sec A = 1 \))
\( = (\sin^2 A + \cos^2 A) + (\csc^2 A + \sec^2 A) + 4 \)
\( = 1 + 1 + \cot^2 A + 1 + \tan^2 A + 4 \) (\( \because 1 + \cot^2 A = \csc^2 A \) & \( 1 + \tan^2 A = \sec^2 A \))
\( = 7 + \tan^2 A + \cot^2 A \)
\( = RHS \)

Question. Prove that: \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A \)
Answer: LHS \( = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} \)
\( = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A) \cos A} = \frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{(1 + \sin A) \cos A} \)
\( = \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{(1 + \sin A) \cos A} = \frac{1 + 1 + 2 \sin A}{(1 + \sin A) \cos A} \)
\( = \frac{2(1 + \sin A)}{(1 + \sin A) \cos A} = \frac{2}{\cos A} = 2 \sec A = RHS \).

Question. Prove that: \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \) 
Answer: LHS \( = \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}} \) (By rationalisation)
\( = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \)
\( = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = RHS \)
Hence Proved.

Question. Prove the following identity, where the angle involved is acute angle for which the expressions are defined. \( \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A \), using the identity \( \csc^2 A = 1 + \cot^2 A \). 
Answer: LHS \( = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \frac{\frac{\cos A - \sin A + 1}{\sin A}}{\frac{\cos A + \sin A - 1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A} \)
\( = \frac{(\cot A + \csc A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1} \) [\( \because \csc^2 A - \cot^2 A = 1 \)]
\( = \frac{(\cot A + \csc A) - [(\csc A + \cot A)(\csc A - \cot A)]}{\cot A - \csc A + 1} \)
\( = \frac{(\csc A + \cot A) (1 - \csc A + \cot A)}{(\cot A - \csc A + 1)} \)
\( = \csc A + \cot A \)
\( = RHS \).

Question. Prove that : \( \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A \) (\( \theta \) is replaced by \( A \))
Answer: To prove: \( \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A \).
Simplifying LHS;
\( \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} \)
\( = \frac{\sin A (1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)} \)
\( = \frac{\sin A}{\cos A} \left[ \frac{1 - (2 \sin^2 A)}{2 \cos^2 A - 1} \right] \)
\( = \frac{\sin A}{\cos A} \left[ \frac{\sin^2 A + \cos^2 A - 2 \sin^2 A}{2 \cos^2 A - (\sin^2 A + \cos^2 A)} \right] \) [\( \because \sin^2 A + \cos^2 A = 1 \)]
\( = \frac{\sin A}{\cos A} \left[ \frac{\cos^2 A - \sin^2 A}{\cos^2 A - \sin^2 A} \right] \)
\( = \frac{\sin A}{\cos A} \times 1 \)
\( = \tan A \) [\( \because \frac{\sin A}{\cos A} = \tan A \)]
LHS = RHS
hence proved. 

Question. Prove that: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta = 1 + \tan \theta + \cot \theta \) 
Answer: LHS \( = \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} \)
\( = \frac{\sin \theta \times \sin \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos \theta \times \cos \theta}{\sin \theta (\cos \theta - \sin \theta)} \)
\( = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} \)
\( = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} = \frac{(\sin \theta - \cos \theta) (\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
\( = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \) ... (i)
\( = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta} \frac{1}{\cos \theta} + 1 \)
\( = \sec \theta \csc \theta + 1 = RHS \)
From (i), \( \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( = \tan \theta + \cot \theta + 1 = RHS \)

Question. Prove that: \( \left( \frac{1 + \tan^2 A}{1 + \cot^2 A} \right) = \left( \frac{1 - \tan A}{1 - \cot A} \right)^2 = \tan^2 A \)
Answer: LHS \( = \left( \frac{1 + \tan^2 A}{1 + \cot^2 A} \right) = \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \)
RHS \( = \left( \frac{1 - \tan A}{1 - \cot A} \right)^2 = \left( \frac{1 - \tan A}{1 - \frac{1}{\tan A}} \right)^2 \)
\( = \left( \frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}} \right)^2 = \left( \frac{1 - \tan A}{\tan A - 1} \times \tan A \right)^2 \)
\( = (- \tan A)^2 = \tan^2 A \)
LHS = RHS.

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. If \( \sin A = \frac{1}{2} \), then the value of \( \cot A \) is 
(a) \( \sqrt{3} \)
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) 1
Answer: (a) \( \sqrt{3} \)

Question. The two legs AB and BC of right triangle ABC are in a ratio 1 : 3. What will be the value of \( \sin C \)?
(a) \( \sqrt{10} \)
(b) \( \frac{1}{\sqrt{10}} \)
(c) \( \frac{3}{\sqrt{10}} \)
(d) \( \frac{1}{2} \)
Answer: (b) \( \frac{1}{\sqrt{10}} \)

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

Question. Which of these is equivalent to \( \frac{2 \tan x (\sec^2 x - 1)}{\cos^3 x} \)? 
(a) \( 2 \tan^3 x \csc x \)
(b) \( 2 \cot^3 x \csc^3 x \)
(c) \( 2 \tan^3 x \sec^3 x \)
(d) \( 2 \cot^3 x \sec^3 x \)
Answer: (c) \( 2 \tan^3 x \sec^3 x \)

Question. If \( 4 \tan \theta = 3 \), then \( \left( \frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta} \right) \) is equal to 
(a) \( \frac{2}{3} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{3}{4} \)
Answer: (c) \( \frac{1}{2} \)

Question. What is the value of \( \frac{3 - \sin^2 60^\circ}{\tan 30^\circ \tan 60^\circ} \)?
(a) \( 2 \frac{1}{4} \)
(b) \( 3 \frac{1}{4} \)
(c) \( 2 \frac{3}{4} \)
(d) \( 3 \frac{3}{4} \)
Answer: (a) \( 2 \frac{1}{4} \)

Question. The value of \( \frac{4 - \sin^2 45^\circ}{\cot k \tan 60^\circ} \) is 3.5. What is the value of \( k \)?
(a) \( 30^\circ \)
(b) \( 45^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Answer: (c) \( 60^\circ \)

Question. The value of \( \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} \) is equal to
(a) \( \cos 60^\circ \)
(b) \( \sin 60^\circ \)
(c) \( \tan 60^\circ \)
(d) \( \sin 30^\circ \)
Answer: (c) \( \tan 60^\circ \)

Question. The value of \( \theta \) for which \( \cos (10^\circ + \theta) = \sin 30^\circ \), is 
(a) \( 50^\circ \)
(b) \( 40^\circ \)
(c) \( 80^\circ \)
(d) \( 20^\circ \)
Answer: (a) \( 50^\circ \)

Question. Given that \( \sin \theta = \frac{a}{b} \), then \( \cos \theta \) is equal to 
(a) \( \frac{b}{\sqrt{b^2 - a^2}} \)
(b) \( \frac{b}{a} \)
(c) \( \frac{\sqrt{b^2 - a^2}}{b} \)
(d) \( \frac{a}{\sqrt{b^2 - a^2}} \)
Answer: (c) \( \frac{\sqrt{b^2 - a^2}}{b} \)

Question. If \( x = r \sin \theta \) and \( y = r \cos \theta \) then the value of \( x^2 + y^2 \) is
(a) \( r \)
(b) \( r^2 \)
(c) \( \frac{1}{r} \)
(d) 1
Answer: (b) \( r^2 \)

Question. If \( \tan x + \sin x = m \) and \( \tan x - \sin x = n \) then \( m^2 - n^2 \) is equal to
(a) \( 4\sqrt{mn} \)
(b) \( \sqrt{mn} \)
(c) \( 2\sqrt{mn} \)
(d) None of the options
Answer: (a) \( 4\sqrt{mn} \)

Question. The value of \( \theta \) for which \( \sin (44^\circ + \theta) = \cos 30^\circ \), is 
(a) \( 46^\circ \)
(b) \( 60^\circ \)
(c) \( 16^\circ \)
(d) \( 90^\circ \)
Answer: (c) \( 16^\circ \)

Very Short Answer Questions

Each of the following questions are of 1 mark.

Question. What is the value of \( \left( \frac{1}{1 + \cot^2 \theta} + \frac{1}{1 + \tan^2 \theta} \right) \)? 
Answer: We have, \( \left( \frac{1}{1 + \cot^2 \theta} + \frac{1}{1 + \tan^2 \theta} \right) \)
\( = \left( \frac{1}{\csc^2 \theta} + \frac{1}{\sec^2 \theta} \right) = \sin^2 \theta + \cos^2 \theta = 1 \)

Question. If \( \tan \alpha = \frac{5}{12} \), find the value of \( \sec \alpha \)? 
Answer: Using identity; \( \sec^2 \alpha - \tan^2 \alpha = 1 \)
\( \sec^2 \alpha = 1 + \tan^2 \alpha \)
\( \implies \) \( \sec^2 \alpha = 1 + \left(\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{144 + 25}{144} \)
\( \implies \) \( \sec^2 \alpha = \frac{169}{144} \)
\( \implies \) \( \sec \alpha = \sqrt{\frac{13^2}{12^2}} = \frac{13}{12} \). 

Question. If \( \sec^2 \theta (1 + \sin \theta)(1 - \sin \theta) = k \), then find the value of \( k \).
Answer: We have, \( \sec^2 \theta (1 + \sin \theta)(1 - \sin \theta) = \sec^2 \theta (1 - \sin^2 \theta) \) (\( \because (a + b)(a - b) = a^2 - b^2 \))
\( = \sec^2 \theta . \cos^2 \theta = 1 \) (\( \because \cos^2 \theta + \sin^2 \theta = 1 \))
\( \therefore k = 1 \)

Question. If \( \sin \theta = \frac{1}{3} \), then find the value of \( 2\cot^2 \theta + 2 \).
Answer: \( 2\cot^2 \theta + 2 = 2(\cot^2 \theta + 1) = 2\csc^2 \theta = \frac{2}{\sin^2 \theta} = \frac{2}{\left(\frac{1}{3}\right)^2} = 2 \times 9 = 18 \)

Question. Evaluate: \( \frac{2 \tan 45^\circ \times \cos 60^\circ}{\sin 30^\circ} \) 
Answer: We have, \[ \frac{2 \tan 45^\circ \times \cos 60^\circ}{\sin 30^\circ} = \frac{2 \times 1 \times \frac{1}{2}}{\frac{1}{2}} = 2 \]

Question. Write the acute angle \( \theta \) satisfying \( \sqrt{3} \sin \theta = \cos \theta \).
Answer: We have, \( \sqrt{3} \sin \theta = \cos \theta \)
\( \implies \) \( \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \tan \theta = \frac{1}{\sqrt{3}} \)
\( \implies \) \( \theta = 30^\circ \)

Question. If \( \sin x + \cos y = 1; x = 30^\circ \) and \( y \) is an acute angle, find the value of \( y \).
Answer: We have, \( \sin x + \cos y = 1 \)
\( \implies \) \( \sin 30^\circ + \cos y = 1 \)
\( \implies \) \( \frac{1}{2} + \cos y = 1 \)
\( \implies \) \( \cos y = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \implies \) \( y = 60^\circ \)

Short Answer Questions-I

Question. Prove that \( 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} = \csc \alpha \). 
Answer: LHS \( = 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} \)
\( = 1 + \frac{\csc^2 \alpha - 1}{1 + \csc \alpha} \)
\( = 1 + \frac{(\csc \alpha - 1)(\csc \alpha + 1)}{(1 + \csc \alpha)} \)
\( = 1 + \csc \alpha - 1 \)
\( = \csc \alpha = RHS \) Proved

Question. Show that \( \tan^4 \theta + \tan^2 \theta = \sec^4 \theta - \sec^2 \theta \). 
Answer: LHS \( = \tan^4 \theta + \tan^2 \theta \)
\( = \tan^2 \theta (\tan^2 \theta + 1) \)
\( = (\sec^2 \theta - 1)(\sec^2 \theta) = \sec^4 \theta - \sec^2 \theta = RHS \) 

Question. Given that \( \sin \theta = \frac{a}{b} \), find the value of \( \tan \theta \).
Answer: \( \sin \theta = \frac{a}{b} \)
\( \implies \) \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b} \)
\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{a/b}{\frac{\sqrt{b^2 - a^2}}{b}} = \frac{a}{\sqrt{b^2 - a^2}} \)

Question. Prove that \( (\sin \alpha + \cos \alpha)(\tan \alpha + \cot \alpha) = \sec \alpha + \csc \alpha \). 
Answer: LHS \( = (\sin \alpha + \cos \alpha)(\tan \alpha + \cot \alpha) \)
\( = (\sin \alpha + \cos \alpha) \left( \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} \right) \)
\( = (\sin \alpha + \cos \alpha) \left( \frac{\sin^2 \alpha + \cos^2 \alpha}{\cos \alpha \sin \alpha} \right) \)
\( = (\sin \alpha + \cos \alpha) \times \frac{1}{\cos \alpha \sin \alpha} \)
\( = \frac{\sin \alpha}{\cos \alpha \sin \alpha} + \frac{\cos \alpha}{\cos \alpha \sin \alpha} \)
\( = \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha} = \sec \alpha + \csc \alpha = RHS \)

Question. If \( \tan \theta = \frac{3}{4} \), find the value of \( \left( \frac{1 - \cos^2 \theta}{1 + \cos^2 \theta} \right) \). 
Answer: Given, \( \tan \theta = \frac{3}{4} \)
Since \( \sec \theta = \sqrt{1 + \tan^2 \theta} \)
\( \implies \) \( \sec \theta = \sqrt{1 + \left( \frac{3}{4} \right)^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{16 + 9}{16}} \)
\( \implies \) \( \sec \theta = \frac{5}{4} \)
\( \implies \) \( \cos \theta = \frac{4}{5} \)
\( \therefore \frac{1 - \cos^2 \theta}{1 + \cos^2 \theta} = \frac{1 - \left( \frac{4}{5} \right)^2}{1 + \left( \frac{4}{5} \right)^2} = \frac{1 - \frac{16}{25}}{1 + \frac{16}{25}} = \frac{25 - 16}{25 + 16} = \frac{9}{41} \)

Question. If \( \tan A = \frac{3}{4} \), find the value of \( \frac{1}{\sin A} + \frac{1}{\cos A} \). 
Answer: \( \tan A = \frac{3}{4} = \frac{3k}{4k} \)
\( \sin A = \frac{3k}{5k} = \frac{3}{5} \), \( \cos A = \frac{4k}{5k} = \frac{4}{5} \)
\( \frac{1}{\sin A} + \frac{1}{\cos A} = \frac{5}{3} + \frac{5}{4} \)
\( = \frac{20 + 15}{12} = \frac{35}{12} \) [CBSE Marking Scheme 2021]

Short Answer Questions-II

Question. If \( \sin \theta + \cos \theta = \sqrt{3} \), then prove that \( \tan \theta + \cot \theta = 1 \). 
Answer: \( \sin \theta + \cos \theta = \sqrt{3} \)
\( \implies \) \( (\sin \theta + \cos \theta)^2 = 3 \)
\( \implies \) \( \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3 \)
\( \implies \) \( 2 \sin \theta \cos \theta = 2 \)
\( \implies \) \( \sin \theta \cos \theta = 1 = \sin^2 \theta + \cos^2 \theta \) (\( \because \sin^2 \theta + \cos^2 \theta = 1 \))
\( \implies \) \( 1 = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta} \)
\( \implies \) \( 1 = \tan \theta + \cot \theta \)
Therefore \( \tan \theta + \cot \theta = 1 \).

Question. Prove that: \( \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = 2 \csc \theta \) 
Answer: LHS \( = \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = \frac{(\sec \theta - 1) + (\sec \theta + 1)}{\sqrt{\sec^2 \theta - 1}} \)
\( = \frac{2 \sec \theta}{\tan \theta} \)
\( = \frac{\frac{2}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = 2 \csc \theta = RHS \) [CBSE Marking Scheme 2019 (30/4/2)]

Question. If \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \), prove that \( \tan \theta = 1 \) or \( \frac{1}{2} \). 
Answer: Given, \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \)
Divide both sides by \( \cos^2 \theta \), we have
\( \frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{3 \sin \theta \cos \theta}{\cos^2 \theta} \)
\( \implies \) \( \sec^2 \theta + \tan^2 \theta = 3 \tan \theta \)
\( \implies \) \( 1 + \tan^2 \theta + \tan^2 \theta = 3 \tan \theta \)
\( \implies \) \( 2 \tan^2 \theta - 3 \tan \theta + 1 = 0 \)
\( \implies \) \( 2 \tan^2 \theta - 2 \tan \theta - \tan \theta + 1 = 0 \)
\( \implies \) \( 2 \tan \theta (\tan \theta - 1) - 1 (\tan \theta - 1) = 0 \)
\( \implies \) \( (\tan \theta - 1)(2 \tan \theta - 1) = 0 \)
\( \implies \) \( \tan \theta - 1 = 0 \) or \( 2 \tan \theta - 1 = 0 \)
\( \implies \) \( \tan \theta = 1 \) or \( 2 \tan \theta = 1 \implies \tan \theta = \frac{1}{2} \)

Question. Prove that: \( (1 + \cot A - \csc A)(1 + \tan A + \sec A) = 2 \) 
Answer: LHS \( = (1 + \cot A - \csc A)(1 + \tan A + \sec A) \)
\( = \left( 1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A} \right) \left( 1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A} \right) \)
\( = \left( \frac{\sin A + \cos A - 1}{\sin A} \right) \left( \frac{\cos A + \sin A + 1}{\cos A} \right) \)
\( = \frac{1}{\sin A \cos A} [(\sin A + \cos A - 1)(\sin A + \cos A + 1)] \)
\( = \frac{1}{\sin A \cos A} [(\sin A + \cos A)^2 - 1] \)
\( = \frac{1}{\sin A \cos A} [\sin^2 A + \cos^2 A + 2 \sin A \cos A - 1] \)
\( = \frac{1}{\sin A \cos A} (1 + 2 \sin A \cos A - 1) \)
\( = \frac{2 \sin A \cos A}{\sin A \cos A} = 2 = RHS \).

Question. Prove that: \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \) 
Answer: LHS \( = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \)
Dividing \( N^r \) and \( D^r \) by \( \cos \theta \)
\( = \frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta} \)
\( = \frac{\tan \theta + \sec \theta - 1}{(\sec^2 \theta - \tan^2 \theta) + \tan \theta - \sec \theta} \)
\( = \frac{\tan \theta + \sec \theta - 1}{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)} \)
\( = \frac{\tan \theta + \sec \theta - 1}{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta - 1)} \)
\( = \frac{1}{\sec \theta - \tan \theta} = RHS \) [CBSE Marking Scheme 2020 (30/4/1)]

Question. If \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \csc \theta = q \), show that \( q(p^2 - 1) = 2p \).
Answer: Given, \( \sin \theta + \cos \theta = p \) ... (i)
Squaring on both sides, we have
\( (\sin \theta + \cos \theta)^2 = p^2 \)
\( \implies \) \( \sin^2 \theta + \cos^2 \theta + 2 \sin \theta . \cos \theta = p^2 \)
\( \implies \) \( 1 + 2 \sin \theta . \cos \theta = p^2 \)
\( \implies \) \( \sin \theta \cos \theta = \frac{p^2 - 1}{2} \) ... (ii)
Also, \( \sec \theta + \csc \theta = q \)
\( \implies \) \( \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q \)
\( \implies \) \( \frac{\sin \theta + \cos \theta}{\sin \theta . \cos \theta} = q \)
\( \implies \) \( \frac{p}{\frac{p^2 - 1}{2}} = q \)
\( \implies \) \( \frac{2p}{p^2 - 1} = q \)
\( \implies \) \( q(p^2 - 1) = 2p \) Proved

Trignometry

∴ AC = 25 cm, BC = (25 – 1)cm = 24 cm

and AB = 7 cm.

For T-ratios of ∠C, we have

base = BC = 24 cm,

perpendicular = AB = 7 cm and

hypotenuse = AC = 25 cm.

∴ sin C =  AB/AC = 7/25

and cos C = BC/AC = 24/25

 

Q.- If sin A = 3/5 , find cos A and tan A.

 

Sol. Since sin A =Perpendicular/Hypotenuse = 3/5, so

We draw a triangle ABC, right angled at B such that

Perpendicular = BC = 3 units,

and, Hypotenuse = AC = 5 units.

By Pythagoras theorem, we have

AC² = AB² + BC²

=> 5² = AB² + 3²

=> AB² = 5² – 3²

=> AB2 = 16 => AB = 4

 

When we consider the t-ratio of ∠A, we have

Base = AB = 4, Perpendicular = BC = 3, Hypotenuse = AC = 5.

∴ cos A = Base/Hypotenuse = 4/5

and, tan A = Perpendicular/Base = 3/4

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