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°. \[\cot \left( {\frac{{ - 15\pi }}{4}} \right) = - \cot \left( {\frac{{15\pi }}{4}} \right)\] \[ = - \cot \left( {4\pi - \frac{\pi }{4}} \right)\] \[ = - \left( { - \cot \frac{\pi }{4}} \right) = 1\]
Exercise 3.1 ⇐
⇒ Exercise 3.3