CBSE Class 10 Maths HOTs Introduction to Trigonometry Set C

VERY SHORT ANSWER TYPE QUESTIONS

Question. In a triangle \(ABC\), write \(\cos \left( \frac{B+C}{2} \right)\) in terms of angle \(A\).
Answer: In a triangle \(A+B+C = 180^{\circ}\) or, \(B+C = 180^{\circ} - A\).
Thus \(\cos \left( \frac{B+C}{2} \right) = \cos \left( \frac{180^{\circ}-A}{2} \right)\)
\(= \cos \left[ 90^{\circ} - \frac{A}{2} \right]\)
\(= \sin \frac{A}{2}\)

Question. If \(\sec \theta \cdot \sin \theta = 0\), then find the value of \(\theta\).
Answer: We have \(\sec \theta \cdot \sin \theta = 0\)
\(\frac{\sin \theta}{\cos \theta} = 0\)
\(\tan \theta = 0 = \tan 0^{\circ}\)
Thus \(\theta = 0^{\circ}\)

Question. If \(A + B = 90^{\circ}\) and \(\sec A = \frac{5}{3}\), then find the value of \(\csc B\).
Answer: We have \(A + B = 90^{\circ}\)
and \(\sec A = \frac{5}{3}\)
or, \(\sec(90^{\circ} - B) = \frac{5}{3}\)
Thus \(\csc B = \frac{5}{3}\)

Question. If \(\tan 2A = \cot(A + 60^{\circ})\), find the value of \(A\) where \(2A\) is an acute angle.
Answer: We have \(\tan 2A = \cot(A + 60^{\circ})\)
\(\cot(90^{\circ} - 2A) = \cot(A + 60^{\circ})\)
\(90^{\circ} - 2A = A + 60^{\circ}\)
\(3A = 30^{\circ}\)
Thus \(A = 10^{\circ}\)

Question. Find the value of \(\frac{\sin 25^{\circ}}{\cos 65^{\circ}} + \frac{\tan 23^{\circ}}{\cot 67^{\circ}}\)
Answer: \(\frac{\sin 25^{\circ}}{\cos 65^{\circ}} + \frac{\tan 23^{\circ}}{\cot 67^{\circ}} = \frac{\sin 25^{\circ}}{\sin(90^{\circ} - 65^{\circ})} + \frac{\tan 23^{\circ}}{\tan(90^{\circ} - 67^{\circ})}\)
\(= \frac{\sin 25^{\circ}}{\sin 25^{\circ}} + \frac{\tan 23^{\circ}}{\tan 23^{\circ}}\)
\(= 1 + 1 = 2\)

Question. If \(\cos 2A = \sin(A - 15)\), find \(A\).
Answer: We have \(\cos 2A = \sin(A - 15^{\circ})\)
\(\sin(90^{\circ} - 2A) = \sin(A - 15^{\circ})\)
\(90^{\circ} - 2A = A - 15^{\circ}\)
\(3A = 105^{\circ}\)
\(A = 35^{\circ}\)

Question. If \(\tan(3x + 30^{\circ}) = 1\) then find the value of \(x\).
Answer: We have \(\tan(3x + 30^{\circ}) = 1 = \tan 45^{\circ}\)
\(3x + 30^{\circ} = 45^{\circ}\)
\(x = 5^{\circ}\)

Question. What happens to value of \(\cos \theta\) when \(\theta\) increases from \(0^{\circ}\) to \(90^{\circ}\).
Answer: \(\cos \theta\) decreases from 1 to 0.

Question. Find the value of \(\tan^2 10^{\circ} - \cot^2 80^{\circ}\).
Answer: We have \(\tan^2 10^{\circ} - \cot^2 80^{\circ} = \tan^2(90^{\circ} - 80^{\circ}) - \cot^2 80^{\circ}\)
[ \(\because \tan(90^{\circ} - \theta) = \cot \theta\) ]
\(= \cot^2 80^{\circ} - \cot^2 80^{\circ}\)
\(= 0\)

Question. If \(A\) and \(B\) are acute angles and \(\sin A = \cos B\), then find the value of \(A + B\).
Answer: We have \(\sin A = \cos B\)
\(\sin A = \sin(90^{\circ} - B)\)
\(A = 90^{\circ} - B\)
\(A + B = 90^{\circ}\)

Question. If \(A\) and \(B\) are acute angles and \(\csc A = \sec B\), then find the value of \(A + B\).
Answer: We have \(\csc A = \sec B\)
\(\csc A = \csc(90^{\circ} - B)\)
\(A = 90^{\circ} - B\)
\(A + B = 90^{\circ}\)

Question. Find the value of \(\cot 10^{\circ} \cot 30^{\circ} \cot 80^{\circ}\).
Answer: \(\cot 10^{\circ} \cot 30^{\circ} \cot 80^{\circ} = \cot(90^{\circ} - 80^{\circ}) \cot 30^{\circ} \cot 80^{\circ}\)
\(= \tan 80^{\circ} \cot 30^{\circ} \cdot \frac{1}{\tan 80^{\circ}}\)
\(= \cot 30^{\circ} = \sqrt{3}\)

Question. Evaluate: \(\frac{\csc 13^{\circ}}{\sec 77^{\circ}} - \frac{\cot 20^{\circ}}{\tan 70^{\circ}}\)
Answer: \(\frac{\csc 13^{\circ}}{\sec 77^{\circ}} - \frac{\cot 20^{\circ}}{\tan 70^{\circ}} = \frac{\csc(90^{\circ} - 77^{\circ})}{\sec 77^{\circ}} - \frac{\cot(90^{\circ} - 70^{\circ})}{\tan 70^{\circ}}\)
\(= \frac{\sec 77^{\circ}}{\sec 77^{\circ}} - \frac{\tan 70^{\circ}}{\tan 70^{\circ}}\)
\(= 1 - 1 = 0\)

Question. Evaluate: \(\frac{\sin 90^{\circ}}{\cos 45^{\circ}} + \frac{1}{\csc 30^{\circ}}\)
Answer: We have \(\frac{\sin 90^{\circ}}{\cos 45^{\circ}} + \frac{1}{\csc 30^{\circ}} = \frac{1}{\frac{1}{\sqrt{2}}} + \frac{1}{2}\)
\(= \sqrt{2} + \frac{1}{2} = \frac{2\sqrt{2} + 1}{2}\)

Question. If \(\sin(36 + \theta)^{\circ} = \cos(16 + \theta)^{\circ}\), then find \(\theta\), where \((36 + \theta)^{\circ}\) and \((16 + \theta)^{\circ}\) are both acute angles.
Answer: We have \(\sin(36 + \theta)^{\circ} = \cos(16 + \theta)^{\circ}\)
\(\cos[90^{\circ} - (36 + \theta)^{\circ}] = \cos(16 + \theta)^{\circ}\)
\(90^{\circ} - 36^{\circ} - \theta = 16^{\circ} + \theta\)
\(2\theta = 90^{\circ} - 36^{\circ} - 16^{\circ} = 38^{\circ}\)
\(\theta = \frac{38^{\circ}}{2} = 19^{\circ}\)

Question. If \(\sqrt{2} \sin \theta = 1\), find the value of \(\sec^2 \theta - \csc^2 \theta\).
Answer: We have \(\sqrt{2} \sin \theta = 1\)
\(\sin \theta = \frac{1}{\sqrt{2}} = \sin 45^{\circ}\)
Thus \(\theta = 45^{\circ}\)
Now \(\sec^2 \theta - \csc^2 \theta = \sec^2 45^{\circ} - \csc^2 45^{\circ}\)
\(= (\sqrt{2})^2 - (\sqrt{2})^2\)
\(= 0\)

Question. Evaluate : \(\frac{\cot (90^{\circ} - \theta) \sin (90^{\circ} - \theta)}{\sin \theta} + \frac{\cot 40^{\circ}}{\tan 50^{\circ}} - (\cos^2 20^{\circ} + \cos^2 70^{\circ})\)
Answer: Given expression : \(\frac{\cot (90^{\circ} - \theta) \sin (90^{\circ} - \theta)}{\sin \theta} + \frac{\cot 40^{\circ}}{\tan 50^{\circ}} - (\cos^2 20^{\circ} + \cos^2 70^{\circ})\)
= \(\frac{\tan \theta \cos \theta}{\sin \theta} + \frac{\cot (90^{\circ} - 50^{\circ})}{\tan 50^{\circ}} - [\cos^2 20^{\circ} + \cos^2 (90^{\circ} - 20^{\circ})]\)
= \(\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} + \frac{\tan 50^{\circ}}{\tan 50^{\circ}} - [\cos^2 20^{\circ} + \sin^2 20^{\circ}]\)
= \(1 + 1 - 1 = 1\)

Question. If \( k + 1 = \sec^2 \theta(1 + \sin \theta)(1 - \sin \theta) \), then find the value of \( k \).
Answer: We have \( k + 1 = \sec^2 \theta(1 + \sin \theta)(1 - \sin \theta) \)
\( = \sec^2 \theta(1 - \sin^2 \theta) \)
\( = \sec^2 \theta \cdot \cos^2 \theta \)
\( = \sec^2 \theta \times \frac{1}{\sec^2 \theta} \)
or, \( k + 1 = 1 \)
or, \( k = 1 - 1 = 0 \)
Thus \( k = 0 \)

Question. Find the value of \( \sin^2 41^\circ + \sin^2 49^\circ \).
Answer: We have
\( \sin^2 41^\circ + \sin^2 49^\circ = \sin^2(90^\circ - 49^\circ) + \sin^2 49^\circ \)
\( = \cos^2 49^\circ + \sin^2 49^\circ \)
\( = 1 \) [\( \cos^2 \theta + \sin^2 \theta = 1 \)]

Question. Express the trigonometric ratio of \( \sec A \) and \( \tan A \) in terms of \( \sin A \).
Answer: We have \( \sec A = \frac{1}{\cos A} = \frac{1}{\sqrt{1 - \sin^2 A}} \)
and \( \tan A = \frac{sin A}{\cos A} = \frac{\sin A}{\sqrt{1 - \sin^2 A}} \)

Question. Prove that : \( \frac{(\sin^4 \theta + \cos^4 \theta)}{1 - 2 \sin^2 \theta \cos^2 \theta} = 1 \)
Answer: \( \frac{(\sin^4 \theta + \cos^4 \theta)}{1 - 2 \sin^2 \theta \cos^2 \theta} = \frac{(\sin^2 \theta)^2 + (\cos^2 \theta)^2}{1 - 2 \sin^2 \theta \cos^2 \theta} \)
\( = \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta}{1 - 2 \sin^2 \theta \cos^2 \theta} \)
\( = \frac{1 - 2 \sin^2 \theta \cos^2 \theta}{1 - 2 \sin^2 \theta \cos^2 \theta} \)
\( = 1 \) Hence Prove

Question. Evaluate the following :
\( \frac{2 \cos^2 60^\circ + 3 \sec^2 30^\circ - 2 \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 45^\circ} \)
Answer: \( \frac{2 \cos^2 60^\circ + 3 \sec^2 30^\circ - 2 \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 45^\circ} \)
\( = \frac{2\left(\frac{1}{2}\right)^2 + 3\left(\frac{2}{\sqrt{3}}\right)^2 - 2(1)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \)
\( = \frac{\frac{2}{4} + 4 - 2}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{5}{2}}{\frac{3}{4}} = \frac{10}{3} \)

SHORT ANSWER TYPE QUESTIONS

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

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: We have \( x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \) ...(1)
and \( x \sin \theta = y \cos \theta \)
or, \( x = \frac{y \cos \theta}{\sin \theta} \) ...(2)
Eliminating \( x \) from eqn. (1) and eqn. (2) we obtain,
\( \frac{y \cos \theta}{\sin \theta} \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta \)
\( y \cos \theta \sin^2 \theta + y \cos^3 \theta = \sin \theta \cos \theta \)
\( y \cos \theta [\sin^2 \theta + \cos^2 \theta] = \sin \theta \cos \theta \)
\( y \cos \theta \times 1 = \sin \theta \cos \theta \)
\( y = \sin \theta \) ...(3)
Substituting this value of \( y \) in eqn. (2) we have,
\( x = \cos \theta \) ...(4)
Squaring and adding eqn. (3) and eqn. (4), we get
\( x^2 + y^2 = \cos^2 \theta + \sin^2 \theta = 1 \) Hence Proved.

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

Question. Evaluate the following : \( \frac{\sec^2 (90^\circ - \theta) - \cot^2 \theta}{2(\sin^2 25^\circ + \sin^2 65^\circ)} - \frac{2 \cos^2 60^\circ \tan^2 28^\circ \tan^2 62^\circ}{3(\sec^2 43^\circ - \cot^2 47^\circ)} \)
Answer: \( \frac{\sec^2 (90^\circ - \theta) - \cot^2 \theta}{2(\sin^2 25^\circ + \sin^2 65^\circ)} - \frac{2 \cos^2 60^\circ \tan^2 28^\circ \tan^2 62^\circ}{3(\sec^2 43^\circ - \cot^2 47^\circ)} \)
\( = \frac{(\text{cosec}^2 \theta - \cot^2 \theta)}{2(\sin^2 25^\circ + \cos^2 25^\circ)} - \frac{2 \times \frac{1}{4} \times \tan^2 28^\circ \times \cot^2 28^\circ}{3(\sec^2 43^\circ - \tan^2 43^\circ)} \)
\( = \frac{1}{2 \times 1} - \frac{\frac{1}{2} \times \tan^2 28^\circ \times \frac{1}{\tan^2 28^\circ}}{3} \)
\( = \frac{1}{2} - \frac{1}{6} = \frac{1}{3} \)

Question. Evaluate : \( \frac{\sec 41^\circ \sin 49^\circ + \cos 29^\circ \text{cosec } 61^\circ - \frac{2}{\sqrt{3}}(\tan 20^\circ \tan 60^\circ \tan 70^\circ)}{3(\sin^2 31^\circ + \sin^2 59^\circ)} \)
Answer: \( \frac{\sec 41^\circ \sin 49^\circ + \cos 29^\circ \text{cosec } 61^\circ - \frac{2}{\sqrt{3}}(\tan 20^\circ \tan 60^\circ \tan 70^\circ)}{3(\sin^2 31^\circ + \sin^2 59^\circ)} \)
\( = \frac{\sec(90^\circ - 49^\circ) \sin 49^\circ + \cos 29^\circ \text{cosec}(90^\circ - 29^\circ) - \frac{2}{\sqrt{3}}[\tan 20^\circ \sqrt{3} \tan(90^\circ - 20^\circ)]}{3[\sin^2 31^\circ + \sin^2(90^\circ - 31^\circ)]} \)
\( = \frac{\text{cosec } 49^\circ \sin 49^\circ + \cos 29^\circ \sec 29^\circ - \frac{2}{\sqrt{3}}[\tan 20^\circ \sqrt{3} \cot 20^\circ]}{3[\sin^2 31^\circ + \cos^2 31^\circ]} \)
\( = \frac{1 + 1 - 2[\tan 20^\circ \cot 20^\circ]}{3[\sin^2 31^\circ + \cos^2 31^\circ]} = \frac{1 + 1 - 2}{3} = \frac{2 - 2}{3} = 0 \)

Question. Evaluate : \( \frac{\cos^2 (45^\circ + \theta) + \cos^2 (45^\circ - \theta)}{\tan (60^\circ + \theta) \tan (30^\circ - \theta)} + \text{cosec } (75^\circ + \theta) - \sec (15^\circ - \theta) \)
Answer: \( \frac{\cos^2 (45^\circ + \theta) + \cos^2 (45^\circ - \theta)}{\tan (60^\circ + \theta) \tan (30^\circ - \theta)} + \text{cosec } (75^\circ + \theta) - \sec (15^\circ - \theta) \)
\( = \frac{\cos^2 (45^\circ + \theta) + \sin^2 (45^\circ + \theta)}{\tan (60^\circ + \theta) \cot (60^\circ + \theta)} + \text{cosec } (75^\circ + \theta) - \text{cosec } (90^\circ - (15^\circ - \theta)) \)
\( = \frac{\cos^2 (45^\circ + \theta) + \sin^2 (45^\circ + \theta)}{\tan (60^\circ + \theta) \cot (60^\circ + \theta)} + \text{cosec } (75^\circ + \theta) - \text{cosec } (75^\circ + \theta) \)
\( = \frac{1}{1} = 1 \)

Question. Find the value of the following without using trigonometric tables : \( \frac{\cos 50^\circ}{2 \sin 40^\circ} + \frac{4(\text{cosec}^2 59^\circ - \tan^2 31^\circ)}{3 \tan^2 45^\circ} - \frac{2}{3} \tan 12^\circ \tan 78^\circ \sin 90^\circ \)
Answer: We have \( \cos 50^\circ = \cos(90^\circ - 40^\circ) = \sin 40^\circ \)
\( \text{cosec}^2 59^\circ = \text{cosec}^2 (90^\circ - 31^\circ) = \sec^2 31^\circ \)
and \( \tan 78^\circ = \tan(90^\circ - 12^\circ) = \cot 12^\circ \)
Hence,
\( \frac{\cos 50^\circ}{2 \sin 40^\circ} + \frac{4(\text{cosec}^2 59^\circ - \tan^2 31^\circ)}{3 \tan^2 45^\circ} - \frac{2}{3} \tan 12^\circ \tan 78^\circ \sin 90^\circ \)
\( = \frac{\sin 40^\circ}{2 \sin 40^\circ} + \frac{4(\sec^2 31^\circ - \tan^2 31^\circ)}{3 \tan^2 45^\circ} - \frac{2}{3} \tan 12^\circ \cot 12^\circ \times 1 \)
\( = \frac{1}{2} + \frac{4}{3} - \frac{2}{3} = \frac{7}{6} \)

Question. Evaluate : \( \frac{\text{cosec}^2 63^\circ + \tan^2 24^\circ}{\cos^2 66^\circ + \sec^2 27^\circ} + \frac{\sin^2 63^\circ + \cos 63^\circ \sin 27^\circ + \sin 27^\circ \sec 63^\circ}{2(\text{cosec}^2 65^\circ - \tan^2 25^\circ)} \)
Answer: \( \frac{\text{cosec}^2 63^\circ + \tan^2 24^\circ}{\cot^2 (90^\circ - 24^\circ) + \text{cosec}^2 (90^\circ - 63^\circ)} + \frac{\sin^2 63^\circ + \cos 63^\circ \sin(90^\circ - 63^\circ) + \sin 27^\circ \sec(90^\circ - 27^\circ)}{2(\text{cosec}^2 65^\circ - \tan^2(90^\circ - 65^\circ))} \)
\( = \frac{\text{cosec}^2 63^\circ + \tan^2 24^\circ}{\tan^2 24^\circ + \text{cosec}^2 63^\circ} + \frac{\sin^2 63^\circ + \cos 63^\circ \cos 63^\circ + \sin 27^\circ \text{cosec } 27^\circ}{2(\text{cosec}^2 65^\circ - \cot^2 65^\circ)} \)
\( = 1 + \frac{\sin^2 63^\circ + \cos^2 63^\circ + \sin 27^\circ \times \frac{1}{\sin 27^\circ}}{2 \times 1} \)
\( = 1 + \frac{1 + 1}{2} = 1 + 1 = 2 \)

Question. If \( \sin \theta + \cos \theta = \sqrt{2} \), then evaluate \( \tan \theta + \cot \theta \).
Answer: We have \( \sin \theta + \cos \theta = \sqrt{2} \)
Squaring both sides, we get
\( (\sin \theta + \cos \theta)^2 = (\sqrt{2})^2 \)
\( \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \)
\( 1 + 2 \sin \theta \cos \theta = 2 \)
\( 2 \sin \theta \cos \theta = 2 - 1 = 1 \)
\( \sin \theta \cos \theta = \frac{1}{2} \)
Now, \( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta} \)
\( = \frac{1}{\cos \theta \sin \theta} = 2 \)

Question. Prove that \( b^2 x^2 - a^2 y^2 = a^2 b^2 \), if :
(1) \( x = a \sec \theta, y = b \tan \theta \), or
(2) \( x = a \text{ cosec } \theta, y = b \cot \theta \)
Answer: (1) We have \( x = a \sec \theta, y = b \tan \theta \),
\( \frac{x^2}{a^2} = \sec^2 \theta, \frac{y^2}{b^2} = \tan^2 \theta \)
or, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = \sec^2 \theta - \tan^2 \theta = 1 \)
Thus \( b^2 x^2 - a^2 y^2 = a^2 b^2 \) Hence Proved
(ii) We have \( x = a \text{ cosec } \theta, y = b \cot \theta \)
\( \frac{x^2}{a^2} = \text{cosec}^2 \theta, \frac{y^2}{b^2} = \cot^2 \theta \)
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = \text{cosec}^2 \theta - \cot^2 \theta = 1 \)
Thus \( b^2 x^2 - a^2 y^2 = a^2 b^2 \) Hence Proved

Question. If \( \text{cosec } \theta - \cot \theta = \sqrt{2} \cot \theta \), then prove that \( \text{cosec } \theta + \cot \theta = \sqrt{2} \text{ cosec } \theta \).
Answer: We have \( \text{cosec } \theta - \cot \theta = \sqrt{2} \cot \theta \)
Squaring both sides we have
\( \text{cosec}^2 \theta + \cot^2 \theta - 2 \text{ cosec } \theta \cot \theta = 2 \cot^2 \theta \)
\( \text{cosec}^2 \theta - \cot^2 \theta = 2 \text{ cosec } \theta \cot \theta \)
\( (\text{cosec } \theta + \cot \theta)(\text{cosec } \theta - \cot \theta) = 2 \text{ cosec } \theta \cot \theta \)
\( (\text{cosec } \theta + \cot \theta)(\sqrt{2} \cot \theta) = 2 \text{ cosec } \theta \cot \theta \)
\( \text{cosec } \theta + \cot \theta = \sqrt{2} \text{ cosec } \theta \)
Hence Proved.

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

Question. Prove that : \(\frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1} = 2 \sec^2 A\)
Answer: \(\frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1}\)
\(= \frac{\csc^2 A + \csc A + \csc^2 A - \csc A}{(\csc A - 1)(\csc A + 1)}\)
\(= \frac{2 \csc^2 A}{\csc^2 A - 1} = \frac{2 \csc^2 A}{\cot^2 A}\)
\(= \frac{2}{\sin^2 A} \times \frac{\sin^2 A}{\cos^2 A}\)
\(= \frac{2}{\cos^2 A} = 2 \sec^2 A\) Hence Proved.

Question. If \(\csc \theta + \cot \theta = p\), then prove that \(\cos \theta = \frac{p^2 - 1}{p^2 + 1}\).
Answer: \(\frac{p^2 - 1}{p^2 + 1} = \frac{(\csc \theta + \cot \theta)^2 - 1}{(\csc \theta + \cot \theta)^2 + 1}\)
\(= \frac{\csc^2 \theta + \cot^2 \theta + 2 \csc \theta \cot \theta - 1}{\csc^2 \theta + \cot^2 \theta + 2 \csc \theta \cot \theta + 1}\)
\(= \frac{1 + \cot^2 \theta + \cot^2 \theta + 2 \csc \theta \cot \theta - 1}{\csc^2 \theta + \csc^2 \theta - 1 + 2 \csc \theta \cot \theta + 1}\)
\(= \frac{2 \cot^2 \theta + 2 \csc \theta \cot \theta}{2 \csc^2 \theta + 2 \csc \theta \cot \theta}\)
\(= \frac{2 \cot \theta (\cot \theta + \csc \theta)}{2 \csc \theta (\csc \theta + \cot \theta)}\)
\(= \frac{\cos \theta}{\sin \theta} \times \sin \theta = \cos \theta\) Hence proved

Question. If \(a \cos \theta + b \sin \theta = m\) and \(a \sin \theta - b \cos \theta = n\), prove that \(m^2 + n^2 = a^2 + b^2\)
Answer: We have
\(m^2 = a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta\) ...(1)
and, \(n^2 = a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta\) ...(2)
Adding equations (1) and (2) we get
\(m^2 + n^2 = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\cos^2 \theta + \sin^2 \theta)\)
\(= a^2(1) + b^2(1)\)
\(= a^2 + b^2\) Hence Proved.

Question. Prove that : \(\frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} = 1 + \sin \theta \cos \theta\).
Answer: \(\frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}\)
\(= \frac{\cos^2 \theta}{1 - \frac{\sin \theta}{\cos \theta}} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}\)
\(= \frac{\cos^3 \theta}{\cos \theta - \sin \theta} - \frac{\sin^3 \theta}{\cos \theta - \sin \theta}\)
\(= \frac{\cos^3 \theta - \sin^3 \theta}{\cos \theta - \sin \theta}\)
\(= \frac{(\cos \theta - \sin \theta)(\cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta)}{(\cos \theta - \sin \theta)}\)
\(= 1 + \sin \theta \cos \theta\) Hence Proved

Question. If \(\cos \theta + \sin \theta = p\) and \(\sec \theta + \csc \theta = q\), prove that \(q(p^2 - 1) = 2p\)
Answer: We have \(\cos \theta + \sin \theta = p\) and \(\sec \theta + \csc \theta = q\)
\(q(p^2 - 1) = (\sec \theta + \csc \theta)[(\cos \theta + \sin \theta)^2 - 1]\)
\(= (\sec \theta + \csc \theta)(\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta - 1)\)
\(= (\sec \theta + \csc \theta)(1 + 2 \sin \theta \cos \theta - 1)\)
\(= (\frac{1}{\cos \theta} + \frac{1}{\sin \theta})(2 \sin \theta \cos \theta)\)
\(= \frac{\sin \theta + \cos \theta}{\cos \theta \sin \theta} \times 2 \sin \theta \cos \theta\)
\(= 2(\sin \theta + \cos \theta) = 2p\) Hence Proved.

Question. If \(x = r \sin A \cos C\), \(y = r \sin A \sin C\) and \(z = r \cos A\), then prove that \(x^2 + y^2 + z^2 = r^2\)
Answer: Since, \(x^2 = r^2 \sin^2 A \cos^2 C\)
\(y^2 = r^2 \sin^2 A \sin^2 C\)
and \(z^2 = r^2 \cos^2 A\)
\(x^2 + y^2 + z^2 = r^2 \sin^2 A \cos^2 C + r^2 \sin^2 A \sin^2 C + r^2 \cos^2 A\)
\(= r^2 \sin^2 A (\cos^2 C + \sin^2 C) + r^2 \cos^2 A\)
\(= r^2 \sin^2 A + r^2 \cos^2 A\)
\(= r^2 (\sin^2 A + \cos^2 A) = r^2\) 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: \(\sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}}\)
\(= \sqrt{\frac{(1 + \sin \theta)}{(1 - \sin \theta)} \times \frac{(1 + \sin \theta)}{(1 + \sin \theta)}} + \sqrt{\frac{(1 - \sin \theta)}{(1 + \sin \theta)} \times \frac{(1 - \sin \theta)}{(1 - \sin \theta)}}\)
\(= \sqrt{\frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta}} + \sqrt{\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}}\)
\(= \sqrt{\frac{(1 + \sin \theta)^2}{\cos^2 \theta}} + \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}}\)
\(= \frac{1 + \sin \theta}{\cos \theta} + \frac{1 - \sin \theta}{\cos \theta}\)
\(= \frac{1 + \sin \theta + 1 - \sin \theta}{\cos \theta}\)
\(= \frac{2}{\cos \theta} = 2 \sec \theta\) Hence Prove

Question. Prove that \((1 - \sin \theta + \cos \theta)^2 = 2(1 + \cos \theta)(1 - \sin \theta)\).
Answer: \((1 - \sin \theta + \cos \theta)^2\)
\(= 1 + \sin^2 \theta + \cos^2 \theta - 2 \sin \theta - 2 \sin \theta \cos \theta + 2 \cos \theta\)
\(= 1 + 1 - 2 \sin \theta - 2 \sin \theta \cos \theta + 2 \cos \theta\)
\(= 2 + 2 \cos \theta - 2 \sin \theta - 2 \sin \theta \cos \theta\)
\(= 2(1 + \cos \theta) - 2 \sin \theta (1 + \cos \theta)\)
\(= (1 + \cos \theta)(2 - 2 \sin \theta)\)
\(= 2(1 + \cos \theta)(1 - \sin \theta)\) Hence Proved

Question. Prove that : \(\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \sec \theta + \tan \theta\)
Answer: \(\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1}\)
\(= \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta - \sec \theta + 1}\)
\(= \frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1}\)
\(= \frac{(\tan \theta + \sec \theta)[1 - \sec \theta + \tan \theta]}{\tan \theta - \sec \theta + 1}\)
\(= \tan \theta + \sec \theta\) Hence Proved

Question. Prove that : \((\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta\)
Answer: \((\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2\)
\(= \sin^2 \theta + \csc^2 \theta + 2 \sin \theta \csc \theta + \cos^2 \theta + \sec^2 \theta + 2 \cos \theta \sec \theta\)
\(= (\sin^2 \theta + \cos^2 \theta) + \csc^2 \theta + 2 \sin \theta \times \frac{1}{\sin \theta} + \sec^2 \theta + 2 \cos \theta \times \frac{1}{\cos \theta}\)
\(= 1 + (1 + \cot^2 \theta) + 2 + (1 + \tan^2 \theta) + 2\)
\(= 7 + \tan^2 \theta + \cot^2 \theta\) Hence Proved

Question. If \(\sin \theta = \frac{c}{\sqrt{c^2 + d^2}}\) and \(d > 0\), find the value of \(\cos \theta\) and \(\tan \theta\).
Answer: We have \(\sin \theta = \frac{c}{\sqrt{c^2 + d^2}}\)
Now \(\cos^2 \theta = 1 - \sin^2 \theta\)
\(= 1 - \left(\frac{c}{\sqrt{c^2 + d^2}}\right)^2\)
\(= 1 - \frac{c^2}{c^2 + d^2}\)
\(= \frac{c^2 + d^2 - c^2}{c^2 + d^2} = \frac{d^2}{c^2 + d^2}\)
Thus \(\cos \theta = \frac{d}{\sqrt{c^2 + d^2}}\)
Again, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{c}{\sqrt{c^2 + d^2}}}{\frac{d}{\sqrt{c^2 + d^2}}}\)
Thus \(\tan \theta = \frac{c}{d}\)

Question. Evaluate : \(\frac{\csc^2 (90^{\circ} - \theta) - \tan^2 \theta}{4(\cos^2 40^{\circ} + \cos^2 50^{\circ})} - \frac{2 \tan^2 30^{\circ} \csc^2 52^{\circ} \sin^2 38^{\circ}}{3(\csc^2 70^{\circ} - \tan^2 20^{\circ})}\)
Answer: \(\csc^2 (90^{\circ} - \theta) = \sec^2 \theta\)
\(\sec^2 \theta - \tan^2 \theta = 1\)
\(\cos^2 40^{\circ} + \cos^2 50^{\circ} = \cos^2 (90^{\circ} - 50^{\circ}) + \cos^2 50^{\circ} = \sin^2 50^{\circ} + \cos^2 50^{\circ} = 1\)
\(\tan^2 30^{\circ} = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\)
\(\csc^2 52^{\circ} \sin^2 38^{\circ} = \csc^2 52^{\circ} \sin^2 (90^{\circ} - 52^{\circ}) = \csc^2 52^{\circ} \cos^2 52^{\circ} = 1\)
and \(\csc^2 70^{\circ} - \tan^2 20^{\circ} = \csc^2 (90^{\circ} - 20^{\circ}) - \tan^2 20^{\circ} = \sec^2 20^{\circ} - \tan^2 20^{\circ} = 1\)
Thus given expression becomes
= \(\frac{1}{4(1)} - \frac{2 \times \frac{1}{3} \times 1}{3(1)}\)
= \(\frac{1}{4} - \frac{2}{9} = \frac{9 - 8}{36} = \frac{1}{36}\)

Question. If \(\sec \theta + \tan \theta = p\), show that \(\sec \theta - \tan \theta = \frac{1}{p}\). Hence, find the values of \(\cos \theta\) and \(\sin \theta\).
Answer: We have \(\sec \theta + \tan \theta = p\)
Now \(\frac{1}{p} = \frac{1}{\sec \theta + \tan \theta} \times \frac{(\sec \theta - \tan \theta)}{\sec \theta - \tan \theta}\)
= \(\frac{\sec \theta - \tan \theta}{\sec^2 \theta - \tan^2 \theta} = \frac{\sec \theta - \tan \theta}{1}\)
= \(\sec \theta - \tan \theta\)
Solving \(\sec \theta + \tan \theta = p\) and \(\sec \theta - \tan \theta = \frac{1}{p}\),
\(2\sec \theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec \theta = \frac{p^2 + 1}{2p} \implies \cos \theta = \frac{2p}{p^2 + 1}\)
and \(2\tan \theta = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \tan \theta = \frac{p^2 - 1}{2p}\)
Thus \(\cos \theta = \frac{2p}{p^2 + 1}\) and \(\sin \theta = \tan \theta \cos \theta = \frac{p^2 - 1}{2p} \cdot \frac{2p}{p^2 + 1} = \frac{p^2 - 1}{p^2 + 1}\)

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

Question. Prove that : \((\sin A + \sec A)^2 + (\cos A + \csc A)^2 = (1 + \sec A \csc A)^2\)
Answer: LHS \( = (\sin A + \sec A)^2 + (\cos A + \csc A)^2\)
= \(\left(\sin A + \frac{1}{\cos A}\right)^2 + \left(\cos A + \frac{1}{\sin A}\right)^2\)
= \(\sin^2 A + \frac{1}{\cos^2 A} + 2 \frac{\sin A}{\cos A} + \cos^2 A + \frac{1}{\sin^2 A} + 2 \frac{\cos A}{\sin A}\)
= \((\sin^2 A + \cos^2 A) + \left(\frac{1}{\sin^2 A} + \frac{1}{\cos^2 A}\right) + 2 \left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right)\)
= \(1 + \frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A} + 2 \left(\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}\right)\)
= \(1 + \frac{1}{\sin^2 A \cos^2 A} + \frac{2}{\sin A \cos A}\)
= \(\left(1 + \frac{1}{\sin A \cos A}\right)^2\)
= \((1 + \sec A \csc A)^2\) Hence Proved.

Question. If \((\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\) Prove that each of the side is equal to \(\pm 1\).
Answer: We have \((\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\)
Multiply both sides by \((\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\)
or, \([(\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)] \times [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)] = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2\)
or, \((\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C) = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2\)
or, \(1 \cdot 1 \cdot 1 = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2\)
or, \((\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) = \pm 1\)

Question. If \(4 \sin \theta = 3\), find the value of \(x\) if \(\sqrt{\frac{\csc^2 \theta - \cot^2 \theta}{\sec^2 \theta - 1}} + 2 \cot \theta = \frac{\sqrt{7}}{x} + \cos \theta\)
Answer: We have \(\sin \theta = \frac{3}{4} \implies \sin^2 \theta = \frac{9}{16}\)
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{9}{16} = \frac{7}{16} \implies \cos \theta = \frac{\sqrt{7}}{4}\)
and \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}\)
Thus \(\sqrt{\frac{\csc^2 \theta - \cot^2 \theta}{\sec^2 \theta - 1}} + 2 \cot \theta = \frac{\sqrt{7}}{x} + \cos \theta\)
\(\sqrt{\frac{1}{\tan^2 \theta}} + 2 \times \frac{\sqrt{7}}{3} = \frac{\sqrt{7}}{x} + \frac{\sqrt{7}}{4}\)
\(\frac{1}{\tan \theta} + \frac{2\sqrt{7}}{3} = \frac{\sqrt{7}}{x} + \frac{\sqrt{7}}{4}\)
\(\frac{\sqrt{7}}{3} + \frac{2\sqrt{7}}{3} - \frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{x}\)
\(\sqrt{7} - \frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{x} \implies \frac{3\sqrt{7}}{4} = \frac{\sqrt{7}}{x} \implies x = \frac{4}{3}\)

HOTS QUESTIONS

Question. If \(\frac{\cos \alpha}{\cos \beta} = m\) and \(\frac{\cos \alpha}{\sin \beta} = n\), show that \((m^2 + n^2) \cos^2 \beta = n^2\).
Answer: We have \(m = \frac{\cos \alpha}{\cos \beta}\) and \(n = \frac{\cos \alpha}{\sin \beta}\)
\(m^2 = \frac{\cos^2 \alpha}{\cos^2 \beta}\) and \(n^2 = \frac{\cos^2 \alpha}{\sin^2 \beta}\)
\((m^2 + n^2) \cos^2 \beta = \left(\frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta}\right) \cos^2 \beta\)
= \(\cos^2 \alpha \left(\frac{1}{\cos^2 \beta} + \frac{1}{\sin^2 \beta}\right) \cos^2 \beta = \cos^2 \alpha \left(\frac{\sin^2 \beta + \cos^2 \beta}{\cos^2 \beta \sin^2 \beta}\right) \cos^2 \beta\)
= \(\cos^2 \alpha \left(\frac{1}{\cos^2 \beta \sin^2 \beta}\right) \cos^2 \beta = \frac{\cos^2 \alpha}{\sin^2 \beta} = n^2\) Hence Proved.

Question. If \(7 \csc \phi - 3 \cot \phi = 7\), prove that \(7 \cot \phi - 3 \csc \phi = 3\).
Answer: We have \(7 \csc \phi - 3 \cot \phi = 7 \implies 7 \csc \phi - 7 = 3 \cot \phi \implies 7(\csc \phi - 1) = 3 \cot \phi\)
\(7(\csc \phi - 1)(\csc \phi + 1) = 3 \cot \phi (\csc \phi + 1) \implies 7(\csc^2 \phi - 1) = 3 \cot \phi (\csc \phi + 1)\)
\(7 \cot^2 \phi = 3 \cot \phi (\csc \phi + 1) \implies 7 \cot \phi = 3(\csc \phi + 1) \implies 7 \cot \phi - 3 \csc \phi = 3\) Hence Proved

Question. Prove that : \(\frac{\cos \theta - \sin \theta + 1}{\cos \theta + \sin \theta - 1} = \csc \theta + \cot \theta\)
Answer: \(\frac{\cos \theta - \sin \theta + 1}{\cos \theta + \sin \theta - 1} = \frac{\sin \theta(\cot \theta - 1 + \csc \theta)}{\sin \theta(\cot \theta + 1 - \csc \theta)}\)
= \(\frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1} = \frac{(\cot \theta + \csc \theta) - (\csc^2 \theta - \cot^2 \theta)}{\cot \theta - \csc \theta + 1}\)
= \(\frac{(\cot \theta + \csc \theta) - (\csc \theta - \cot \theta)(\csc \theta + \cot \theta)}{\cot \theta - \csc \theta + 1}\)
= \(\frac{(\csc \theta + \cot \theta)[1 - (\csc \theta - \cot \theta)]}{\cot \theta - \csc \theta + 1} = \frac{(\csc \theta + \cot \theta)(1 - \csc \theta + \cot \theta)}{\cot \theta - \csc \theta + 1}\)
= \(\csc \theta + \cot \theta\) Hence Proved