Class 9: Trigonometric Ratios – Exercise 18.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Evaluate each of the following:

Answer:

$\displaystyle \text{i) } \mathrm{cosec} 30^o + \cot 45^o = 2 + 1 = 3$

$\displaystyle \text{ii) }\cos 30^o \cos 45^o - \sin 30^o \sin 45^o = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3} -1}{2\sqrt{2}}$

$\displaystyle \text{iii) }\tan 30^o \sec 45^o + \tan 60^o \sec 30^o = \frac{1}{\sqrt{3}} \times \sqrt{2} + \sqrt{3} \times \frac{2}{\sqrt{3}} = \frac{\sqrt{2} + 2\sqrt{3}}{\sqrt{3}}$

$\displaystyle \text{iv) }\sin 30^o \sin 45^o + \cos 30^o \cos 45^o = \frac{1}{2} \times \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

$\displaystyle \text{v) }\frac{\sin^2 45^o + \cos^2 45^o}{\tan^2 60^o} = \{ \frac{ (\frac{1}{\sqrt{2}} )^2+ ( \frac{1}{\sqrt{2}} )^2}{(\sqrt{3})^2} \} = \frac{\frac{1}{2}+\frac{1}{2}}{3} = \frac{1}{3}$

$\displaystyle \text{vi) }\frac{\sin 30^o - \sin 90^o 2 \cos 0^o}{ \tan 30^o \tan 60^o} = \{ \frac{\frac{1}{2} - 1 + 2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}} \} = \frac{3}{2}$

$\displaystyle \text{vii) }\frac{\sin 60^o}{\cos^2 45^o} - \cot 30^o + 15 \cos 90^o = \{ \frac{\frac{\sqrt{3}}{2}}{( \frac{1}{\sqrt{2}} )^2} \} - \sqrt{3} + 15 \times 0 = \sqrt{3} - \sqrt{3} = 0$

$\displaystyle \text{viii) }\frac{5 \sin^2 30^o + \cos^2 45^o - 4 \tan^2 30^o}{2 \sin 30^o \cos 30^o + \tan 45^o} = \{ \frac{5 \times (\frac{1}{2} )^2 + ( \frac{1}{\sqrt{2}} )^2 - 4 ( \frac{1}{\sqrt{3}} )^2}{ 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} + 1} \}$

$\displaystyle = \frac{\frac{5}{4} + \frac{1}{2} - \frac{4}{3}}{\frac{\sqrt{3}}{2} + 1} = \frac{5}{6} (2 - \sqrt{3})$

$\displaystyle \\$

Question 2: Find the value of \theta in each of the following:

Answer:

$\displaystyle \text{i) }2 \sin \theta =\sqrt{3}$

$\displaystyle \Rightarrow \sin 2 \theta = \frac{\sqrt{3}}{2} \Rightarrow \sin 2 \theta = \sin 60^o$

$\displaystyle \Rightarrow 2 \theta = 60^o \Rightarrow \theta = 30^o$

$\displaystyle \text{ii) }2 \cos 3 \theta = 1$

$\displaystyle \Rightarrow \cos 3\theta = \cos 60^o \Rightarrow 3 \theta = 60^o$

$\displaystyle \Rightarrow \theta = 20^o$

$\displaystyle \text{iii) }\sqrt{3} \tan 2\theta - 3 = 0$

$\displaystyle \Rightarrow \tan 2 \theta = \frac{3}{\sqrt{3}} = \sqrt{3} \Rightarrow \tan 3 \theta = \tan 60^o$

$\displaystyle \Rightarrow 2 \theta = 60^o \Rightarrow \theta = 30^o$

$\displaystyle \\$

Question 3: Evaluate / Prove the following:

Answer:

$\displaystyle \text{i) }\frac{\cos 37^o}{\sin 53^o} = \frac{\cos (90^o-53^o)}{\sin 53^o} = \frac{\sin 53^o}{\sin 53^o} = 1$

$\displaystyle \text{ii) }\frac{\tan 54^o}{\cot 36^o} = \frac{\tan (90^o-36^o)}{\cot 36^o} = \frac{\cot 36^o}{\cot 36^o} = 1$

$\displaystyle \text{iii) }\sin 39^o - \cos 51^o = \sin 39^o - \cos (90^o-39^o) = \sin 39^o - \sin 39^o = 0$

$\displaystyle \text{iv) }\cot 34^o - \tan 56^o = \cot 34^o - \tan (90^o-34^o) = \cot 34^o - \cot 34^o = 0$

$\displaystyle \text{v) }\frac{\cos 80^o}{\sin 10^o} + \cos 59^o \mathrm{cosec} 31^o = \frac{\cos (90^o-10^o)}{\sin 10^o} + \cos 59^o \mathrm{cosec} (90^o-59^o)$

$\displaystyle = \frac{\sin 10^o}{\sin 10^o} + \cos 59^o \sec 59^o = 1 + 1 = 2$

$\displaystyle \text{vi) }\sec 50^o \sin 40^o + \cos 40^o \mathrm{cosec} 50^o$

$\displaystyle = \sec 50^o \sin (90^o - 50^o) + \cos 40^o \mathrm{cosec} (90^o-40^o)$

$\displaystyle = \sec 50^o \cos 50^o + \cos 40^o \sec 40^o = 1 + 1 = 2$

$\displaystyle \text{vii) }\Big( \frac{\sin 35^o}{\cos 55^o} \Big)^2 + \Big( \frac{\cos 55^o}{\sin 35^o} \Big)^2 - 2 \cos 60^o$

$\displaystyle = \Big( \frac{\sin (90^o-55^o)}{\cos 55^o} \Big)^2 + \Big( \frac{\cos (90^o-35^o)}{\sin 35^o} \Big)^2 - 2 \cos 60^o$

$\displaystyle = \Big( \frac{\cos 55^o}{\cos 55^o} \Big)^2 + \Big( \frac{\sin 35^o}{\sin 35^o} \Big)^2 - 2 \times \frac{1}{2}$

$\displaystyle = 1 + 1 - 1 = 1$

$\displaystyle \text{vii) }\cos (40^o - \theta ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \cos^2 50^o}{\sin^2 40^o + \sin^2 50^o}$

$\displaystyle = \sin (90 - (40^o - \theta) ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \cos^2 (90-40^o)}{\sin^2 40^o + \sin^2 (90-40^o)}$

$\displaystyle = \sin (50^o + \theta ) - \sin ( 50^o + \theta) + \frac{\cos^2 40^o + \sin^2 40^o}{\sin^2 40^o + \cos^2 40^o}$

$\displaystyle = 0 + 1 = 1$

$\displaystyle \text{viii) }\cot 12^o \cot 38^o \cot 52^o \cot 60^o \cot 78^o$

$\displaystyle = \cot (90^o-78^o) \cot (90^o-52^o) \cot 52^o \cot 60^o \cot 78^o$

$\displaystyle = \tan 78^o \tan 52^o \cot 52^o \cot 60^o \cot 78^o$

$\displaystyle = \cot 60^o = \frac{1}{\sqrt{3}}$

$\displaystyle \text{ix) } \tan 5^o \tan 25^o \tan 30^o \tan 65^o \tan 85^o$

$\displaystyle = \tan (90^o-85^o) \tan (90^o-65^o) \tan 30^o \tan 65^o \tan 85^o$

$\displaystyle = \cot 85^o \cot 65^o \tan 30^o \tan 65^o \tan 85^o$

$\displaystyle = \tan 30^o = \frac{1}{\sqrt{3}}$

$\displaystyle \\$

Question 4: Express each of the following in terms of trigonometric ratios of angles between $\displaystyle 0$ and $\displaystyle 45^o$ .

Answer:

$\displaystyle \text{i) }\sin 85^o + \mathrm{cosec} 85^o = \sin (90^o-5^o) + \mathrm{cosec} (90^o-5^o) = \cos 5^o + \sec 5^o$

$\displaystyle \text{ii) } \mathrm{cosec} 69^o + \cot 69^o = \mathrm{cosec} (90^o-21^o) + \cot (90^o-21^o) = \sec 21^o + \tan 21^o$

$\displaystyle \\$

Question 5: Prove that:

$\displaystyle \text{i) }\tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o = 1$

$\displaystyle \text{ii) }\cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o = 0$

Answer:

i) $\displaystyle \text{LHS} = \tan 1^o \tan 2^o \tan 3^o \cdots \tan 89^o$

$\displaystyle = \tan (90^o-89^o) \tan (90^o-88^o) \tan (90^o-87^o) \cdots \tan 87^o \tan 88^o \tan 89^o$

$\displaystyle = \cot 89^o \cot 88^o \cot 87^o \cdots \tan 87^o \tan 88^o \tan 89^o$

$\displaystyle = 1 \times 1 \times 1 \cdots \times 1 = 1 = RHS$

Hence Proved.

ii) $\displaystyle \text{LHS} = \cos 1^o \cos 2^o \cos 3^o \cdots \cos 180^o$

$\displaystyle = \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o \cos 90^o \cos 91^o \cdots \cos 179^o \cos 180^o$

$\displaystyle = \cos 1^o \cos 2^o \cos 3^o \cdots \cos 89^o (0) \cos 91^o \cdots \cos 179^o \cos 180^o$

$\displaystyle = 0 =$ RHS.

Hence Proved.

$\displaystyle \\$

Question 6: If $\displaystyle A + B = 90^o$ , prove that:

$\displaystyle \sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} } = \tan A$

Answer:

Given $\displaystyle A + B = 90^o \Rightarrow B = 90^o - A$

$\displaystyle \text{LHS} = \sqrt{ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} }$

$\displaystyle = \sqrt{ \frac{\tan A \tan (90^o - A) + \tan A \cot (90^o - A)}{\sin A \sec (90^o - A)} - \frac{\sin^2 (90^o - A)}{\cos^2 A} }$

$\displaystyle = \sqrt{ \frac{\tan A \cot A + \tan^2 A }{\sin A \mathrm{cosec} A} - \frac{\cos^2 A}{\cos^2 A} }$

$\displaystyle = \sqrt{1 + \tan^2 A - 1} = \sqrt{\tan^2 A} = \tan A =$ RHS

Hence Proved.

$\displaystyle \\$

Question 7: If $\displaystyle A, B$ and $\displaystyle C$ are interior angles of triangle ABC, prove that

$\displaystyle \tan \big( \frac{B+C}{2} \Big) = \cot \Big( \frac{A}{2} \Big)$

Answer:

Given $\displaystyle A + B + C = 180^o$

$\displaystyle \Rightarrow B + C = 180^o - A$

$\displaystyle \Rightarrow \big( \frac{B+C}{2} \Big) = \Big( 90^o - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \tan \big( \frac{B+C}{2} \Big) = \tan \Big( 90^o - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \tan \big( \frac{B+C}{2} \Big) = \cot \Big( \frac{A}{2} \Big)$

$\displaystyle \\$

Question 8: Find $\displaystyle \theta$ if $\displaystyle \sin (\theta + 36^o) = \cos \theta$ when $\displaystyle \theta + 36^o$ is an acute angle.

Answer:

Given $\displaystyle \sin (\theta + 36^o) = \cos \theta$

$\displaystyle \Rightarrow \cos [90^o - (\theta + 36^o) ] = \cos \theta$

$\displaystyle \Rightarrow [ (\theta + 36^o) ] = \theta$

$\displaystyle \Rightarrow 2 \theta = 54^o$

$\displaystyle \Rightarrow \theta = 27^o$

$\displaystyle \\$

Question 9: If $\displaystyle \tan 2 \theta = \cot (\theta + 6^o)$ , when $\displaystyle 2 \theta$ and $\displaystyle \theta+6^o$ are acute angles, find the value of $\displaystyle \theta$ .

Answer:

Given $\displaystyle \tan 2 \theta = \cot (\theta + 6^o)$

$\displaystyle \Rightarrow \cot (90^o- 2 \theta) = \cot (\theta+6^o)$

$\displaystyle \Rightarrow 90^o - 2 \theta = \theta + 6^o$

$\displaystyle \Rightarrow 84^o = 3 \theta$

$\displaystyle \Rightarrow \theta = 28^o$

$\displaystyle \\$

Question 10: If $\displaystyle A, B$ and $\displaystyle C$ are interior angles of $\displaystyle \triangle ABC$ , show that

$\displaystyle \text{i) }\sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)$ $\displaystyle \text{ii) }\cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)$

Answer:

i) Given $\displaystyle A + B + C = 180^o$

$\displaystyle \Rightarrow B + C = 180^o - A$

$\displaystyle \Rightarrow \Big( \frac{B+C}{2} \Big) = \Big( 90^o - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \sin \Big(90^o - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)$

ii) Similarly,

$\displaystyle \Rightarrow \cos \Big( \frac{B+C}{2} \Big) = \cos \Big( 90^o - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)$

$\displaystyle \\$

Question 11: Find the value of $\displaystyle \theta$ for

$\displaystyle \text{i) }\cos 2\theta = \sin 4\theta$ where $\displaystyle 2\theta, 4\theta < 90^o$

$\displaystyle \text{ii) }\sin 3\theta = \cos (\theta - 6^o)$ where $\displaystyle 3\theta, (\theta-6^o) < 90^o$

Answer:

$\displaystyle \text{i) }\cos 2\theta = \sin 4\theta$

$\displaystyle \Rightarrow \cos 2\theta = \cos (90^o - 4\theta)$

$\displaystyle \Rightarrow 2\theta = 90^o- 4\theta$

$\displaystyle \Rightarrow 6\theta = 90^o$

$\displaystyle \Rightarrow \theta = 15^o$

$\displaystyle \text{ii) }\sin 3\theta = \cos (\theta - 6^o)$

$\displaystyle \Rightarrow \sin 3\theta = \sin [90^o - (\theta-6)]$

$\displaystyle \Rightarrow 3\theta = 90^o - (\theta - 6^o)$

$\displaystyle \Rightarrow 3\theta = 90^o - \theta +6^o$

$\displaystyle \Rightarrow 4\theta = 96^o \Rightarrow \theta = 19^o$

$\displaystyle \\$

Question 12: Prove the following:

$\displaystyle \text{i) }\tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o = 1$

$\displaystyle \text{ii) }\frac{\cos 80^o}{\sin 10^o} + \cos 59^o \mathrm{cosec} 31^o = 2$

$\displaystyle \text{iii) }\frac{\cos (90^o - A) \sec (90^o-A) \tan A}{ \mathrm{cosec} (90^o - A) \sin (90^o - A) \cot (90^o- A)} + \frac{\tan (90^o-A)}{\cot A} = 2$

Answer:

$\displaystyle \text{i) LHS} = \tan 20^o \tan 35^o \tan 45^o \tan 55^o \tan 70^o$

$\displaystyle = \tan (90^o-70^o) \tan (90^o-55^o) \tan 45^o \tan 55^o \tan 70^o$

$\displaystyle = \cot 70^o \cot 55^o \tan 45^o \tan 55^o \tan 70^o$

$\displaystyle = \tan 45^o = 1 =$ RHS

Hence Proved.

$\displaystyle \text{ii) LHS} = \frac{\cos 80^o}{\sin 10^o} + \cos 59^o \mathrm{cosec} 31^o$

$\displaystyle = \frac{\cos (90^o-10^o)}{\sin 10^o} + \cos 59^o \mathrm{cosec} (90^o- 59^o)$

$\displaystyle = \frac{\sin 10^o}{\sin 10^o} + \cos 59^o \sec 59^o$

$\displaystyle = 1 + 1 = 2 =$ RHS

Hence Proved.

$\displaystyle \text{iii) LHS} = \frac{\cos (90^o - A) \sec (90^o-A) \tan A}{ \mathrm{cosec} (90^o - A) \sin (90^o - A) \cot (90^o- A)} + \frac{\tan (90^o-A)}{\cot A}$

$\displaystyle = \frac{\sin A \mathrm{cosec} A \tan A}{\sec A \cos A \tan A} + \frac{\cot A}{ \cot A}$

$\displaystyle = 1 + 1 = 2$

$\displaystyle \\$

$\displaystyle \text{Question 13: What is the max value of } \text{i) }\frac{1}{\sec A} \text{ and } \text{ii) }\frac{1}{ \mathrm{cosec} A}$

Answer:

$\displaystyle \text{i) }\frac{1}{\sec A} = \cos A \Rightarrow \text{ max value } = 1$

$\displaystyle \text{ii) }\frac{1}{ \mathrm{cosec} A} = \sin A \Rightarrow \text{ max value } = 1$

$\displaystyle \\$

$\displaystyle \text{Question 14: If } \tan A = \frac{4}{5} \text{ , find the value of } \frac{\cos A - \sin A}{\cos A + \sin A}$

Answer:

$\displaystyle \frac{\cos A - \sin A}{\cos A + \sin A} = \frac{1 - \tan A}{1 + \tan A} = \frac{1 - \frac{4}{5}}{1 + \frac{4}{5}} = \frac{1}{9}$

$\displaystyle \\$

$\displaystyle \text{Question 15: If } \tan A = \frac{1}{\sqrt{5}} \text{ , what is the value of } \frac{ \mathrm{cosec}^2 A - \sec^2 A}{ \mathrm{cosec}^2 A + \sec^2 A}$