11th NCERT/CBSE Trigonometric Functions Exercise 3.2 Questions 10
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### Find the value of other five trignometric functions in Exercises 1 to 5

Question (1)

$\cos x = - \frac{1}{2}\text{, x lies in third quadrant}$

Solution

$\cos x = - \frac{1}{2}$ $\therefore \sec x = \frac{1}{{\cos x}} = \frac{1}{{\left( { - \frac{1}{2}} \right)}} = - 2$ ${\sin ^2}x + {\cos ^2}x = 1$ $\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$ ${\sin ^2}x = 1 - {\left( { - \frac{1}{2}} \right)^2}$ ${\sin ^2}x = 1 - \frac{1}{4} = \frac{3}{4}$ $\sin x = \pm \frac{{\sqrt 3 }}{2}$ since x lies in the 2nd quadrant, the value of sin x will be negative $\therefore \sin x = - \frac{{\sqrt 3 }}{2}$ $\therefore \text{cosec x }= \frac{1}{{\sin x}} = \frac{1}{{\left( { - \frac{{\sqrt 3 }}{2}} \right)}} = - \frac{2}{{\sqrt 3 }}$ $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\left( { - \frac{{\sqrt 3 }}{2}} \right)}}{{\left( { - \frac{1}{2}} \right)}} = \sqrt 3$ $\cot x = \frac{1}{{\tan x}} = \frac{1}{{\sqrt 3 }}$

Question (2)

$\sin x = \frac{3}{5}\text{, x lies in second quadrant}$

Solution

$\sin x = \frac{3}{5}$ $\cosec x = \frac{1}{{\sin x}} = \frac{1}{{\left( {\frac{3}{5}} \right)}} = \frac{5}{3}$ ${\sin ^2}x + {\cos ^2}x = 1$ $\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$ ${\cos ^2}x = 1 - {\left( {\frac{3}{5}} \right)^2}$ ${\cos ^2}x = 1 - \frac{9}{{25}} = \frac{{16}}{{25}}$ $\cos x = \pm \frac{4}{5}$ Since x lies in the 2nd quadrant, the value of cos x will be negative
$\cos x = - \frac{4}{5}$ $\sec x = \frac{1}{{\cos x}} = \frac{1}{{\left( { - \frac{4}{5}} \right)}} = - \frac{5}{4}$ $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\left( {\frac{3}{5}} \right)}}{{\left( { - \frac{4}{5}} \right)}} = - \frac{3}{4}$ $\cot x = \frac{1}{{\tan x}} = - \frac{4}{3}$

Question (3)

$\cot x = \frac{3}{4}\text{, x lies in third quadrant}$

Solution

$\cot x = \frac{3}{4}$ $\tan x = \frac{1}{{\cot x}} = \frac{1}{{\left( {\frac{3}{4}} \right)}} = \frac{4}{3}$ $1 + {\tan ^2}x = {\sec ^2}x$ $\Rightarrow 1 + {\left( {\frac{4}{3}} \right)^2} = {\sec ^2}x$ $\Rightarrow 1 + \frac{{16}}{9} = {\sec ^2}x$ $\Rightarrow \frac{{25}}{9} = {\sec ^2}x$ $\sec x = \pm \frac{5}{3}$ Since x lies in the 3rd quadrant, the value of sec x will be negative
$\sec x = - \frac{5}{3}$ $\cos x = \frac{1}{{\sec x}} = \frac{1}{{ - \frac{5}{3}}} = - \frac{3}{5}$ $\tan x = \frac{{\sin x}}{{\cos x}}$ $\Rightarrow \frac{4}{3} = \frac{{\sin x}}{{\left( {\frac{{ - 3}}{5}} \right)}}$ $\Rightarrow \sin x = \left( {\frac{4}{3}} \right) \times \left( {\frac{{ - 3}}{5}} \right) = - \frac{4}{5}$ $\cosecx = \frac{1}{{\sin x}} = - \frac{5}{4}$

Question (4)

$\sec x = \frac{{13}}{5}\text{, x lies in fourth quadrant}$

Solution

$\sec x = \frac{{13}}{5}$ $\cos x = \frac{1}{{\sec x}} = \frac{1}{{\frac{{13}}{5}}} = \frac{5}{{13}}$ ${\sin ^2}x + {\cos ^2}x = 1$ $\Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$ $\Rightarrow {\sin ^2}x = 1 - {\left( {\frac{5}{{12}}} \right)^2}$ $\Rightarrow {\sin ^2}x = 1 - \frac{{25}}{{169}} = \frac{{144}}{{169}}$ $\Rightarrow \sin x = \pm \frac{{12}}{{13}}$ Since x lies in the 4th quadrent, the value off sin x will be negative
$\therefore \sin x = - \frac{{12}}{{13}}$ $\cosec x = \frac{1}{{\sin x}} = \frac{1}{{\left( { - \frac{{12}}{{13}}} \right)}} = - \frac{{13}}{{12}}$ $\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{\left( { - \frac{{12}}{{13}}} \right)}}{{\left( {\frac{5}{{13}}} \right)}} = - \frac{{12}}{5}$ $\cot x = \frac{1}{{\tan x}} = \frac{1}{{\left( { - \frac{{12}}{5}} \right)}} = - \frac{5}{{12}}$

Question (5)

$\tan x = - \frac{5}{{12}}\text{, x lies in second quadrant}$

Solution

$\tan x = - \frac{5}{{12}}$ $\cot x = \frac{1}{{\tan x}} = \frac{1}{{\left( { - \frac{5}{{12}}} \right)}} = - \frac{{12}}{5}$ $1 + {\tan ^2}x = {\sec ^2}x$ $\Rightarrow 1 + {\left( { - \frac{5}{{12}}} \right)^2} = {\sec ^2}x$ $\Rightarrow 1 + \frac{{25}}{{144}} = {\sec ^2}x$ $\Rightarrow \frac{{169}}{{144}} = {\sec ^2}x$ $\Rightarrow \sec x = \pm \frac{{13}}{{12}}$ Since x lies in the 2nd quadrant, the value of sec x will be negative. $\sec x = - \frac{{13}}{{12}}$ $\cos x = \frac{1}{{\sec x}} = \frac{1}{{\left( { - \frac{{13}}{{12}}} \right)}} = - \frac{{12}}{{13}}$ $\tan x = \frac{{\sin x}}{{\cos x}}$ $\Rightarrow - \frac{5}{{12}} = \frac{{\sin x}}{{\left( { - \frac{{12}}{{13}}} \right)}}$ $\Rightarrow \sin x = \left( { - \frac{5}{{12}}} \right) \times \left( { - \frac{{12}}{{13}}} \right) = \frac{5}{{13}}$ $\text{cosec} x = \frac{1}{{\sin x}} = \frac{1}{{\left( {\frac{5}{{13}}} \right)}} = \frac{{13}}{5}$

### Find the value of the trigonometric function in Exercises 6 to 10

Question (6)

sin 765°

Solution

It is known that the values of sin x repeat after an interval of 2π or 360°. $\sin {765^o} = \sin \left( {2 \times {{360}^o} + {{45}^o}} \right) = \sin {45^o} = \frac{1}{{\sqrt 2 }}$

Question (7)

cosec (-1410:°)

Solution

It is known that the values of cosec x repeat after an interval of 2π or 360°. $\cos ec\left( { - 1410} \right) = - \cos ec\left( {1410} \right)$ $= - \cos ec\left( {8\pi - 30} \right)$ $= - \left[ { - \cos ec30} \right]$ $= \cos ec30 = 2$

Question (8)

$\tan \frac{{19\pi }}{3}$

Solution

It is known that the values of tan x repeat after an interval of π or 180°. $\tan \frac{{19\pi }}{3} = \tan 6\frac{1}{3}\pi = \tan \left( {6\pi + \frac{\pi }{3}} \right) = \tan \frac{\pi }{3} = \tan {60^0} = \sqrt 3$

Question (9)

$\sin \left( { - \frac{{11\pi }}{3}} \right)$

Solution

It is known that the values of sin x repeat after an interval of 2π or 360°. $\sin \left( { - \frac{{11\pi }}{3}} \right) = - \sin \left( {\frac{{11\pi }}{3}} \right)$ $= - \sin \left( {4\pi - \frac{\pi }{3}} \right)$ $= - \left[ { - \sin \frac{\pi }{3}} \right] = \frac{{\sqrt 3 }}{2}$

Question (10)

$\cot \left( { - \frac{{15\pi }}{4}} \right)$

Solution

It is known that the values of cot x repeat after an interval of π or 180°. $\cot \left( {\frac{{ - 15\pi }}{4}} \right) = - \cot \left( {\frac{{15\pi }}{4}} \right)$ $= - \cot \left( {4\pi - \frac{\pi }{4}} \right)$ $= - \left( { - \cot \frac{\pi }{4}} \right) = 1$