Ex 3.2 Trigonomety ncert maths solution class 11

  EXERCISE 3.2(Trigonometric Function)

Find the values of other five trigonometric functions in Exercises 1 to 5.(Ex 3.2 Trigonomety ncert maths solution class 11)

Question 1: \cos x = -\frac{1}{2}, x lies in third quadrant.

Solution: Given, \cos x = -\frac{1}{2}

Since, \sec x = \frac{1}{\cos x}

\Rightarrow \sec x = \frac{1}{-1/2}

\Rightarrow \sec x = -2

We know that

\sin^2 x+\cos^2x=1

\Rightarrow \sin^2 x = 1-\cos^2x

\Rightarrow \sin^2x=1-(-\frac{1}{2})^2

\Rightarrow \sin^2x = 1-\frac{1}{4}=\frac{3}{4}

\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}

Since x lies in third quadrant

\sin x = -\frac{\sqrt{3}}{2}

We know that

\operatorname{cosec x}=\frac{1}{\sin x}

\Rightarrow \operatorname{cosec x}= \frac{1}{-\sqrt{3}/2}=-\frac{2}{\sqrt{3}}

Since, \tan x = \frac{\sin x}{\cos x}

\Rightarrow \tan x = \frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}

Now \cot x = \frac{1}{\tan x}

\Rightarrow \cot x = \frac{1}{\sqrt{3}}

Question 2: \sin x = \frac{3}{5}, x lies in second quadrant

Solution: Given, \sin x = \frac{3}{5}

Since,\operatorname{cosec x} =\frac{1}{\sin x}= \frac{1}{3/5}

\Rightarrow \operatorname{cosec x}=\frac{5}{3}

We know that

\cos^2 x = 1-\sin^2 x

\Rightarrow \cos^2 x = 1-(\frac{3}{5})^2 = 1-\frac{9}{25}

\Rightarrow \cos^2 x = \frac{16}{25}

\Rightarrow \cos x = =\pm \frac{4}{5}

X lies in second quadrant

\cos x = -\frac{4}{5}

Since, \sec x =\frac{1}{\cos x}=\frac{1}{-4/5}

\Rightarrow \sec x = -\frac{5}{4}

Again we know that

\tan x = \frac{\sin x}{\cos x}= \frac{3/5}{-4/5}

\Rightarrow \tan x = -\frac{3}{4}

And \cot x = \frac{1}{\tan x}=\frac{1}{-3/4}

\Rightarrow \cot x = -\frac{4}{3}

Question 3: \cot x = \frac{3}{4}, x lies in third quadrant.

Solution: Given \cot x =\frac{3}{4}

Since, \tan x = \frac{1}{\cot x}

\Rightarrow \tan x = \frac{1}{3/4}=\frac{4}{3}

We know that

\sec^2x = 1+\tan^2x

\Rightarrow \sec^2x = 1+(\frac{4}{3})^2

= 1+\frac{16}{9}= \frac{25}{9}

\Rightarrow \sec x = \pm \frac{5}{3}

x lies in third quadrant

\sec x = -\frac{5}{3}

Now, \cos x = \frac{1}{\sec x}

\Rightarrow \cos x = \frac{1}{-5/3}=-\frac{3}{5}

Again we know that

\sin^2x = 1-\cos^2x

\Rightarrow \sin^2x = 1-(-\frac{3}{5})^2

= 1-\frac{9}{25}

\Rightarrow \sin^2x= \frac{16}{25}

\Rightarrow \sin x = \pm \frac{4}{5}

x lies in third quadrant

\sin x = -\frac{4}{5}

Since, \operatorname{cosec x}=\frac{1}{\sin x}

= \frac{1}{-4/5}= -\frac{5}{4}

Question 4: \sec x = \frac{13}{5}, x lies in fourth quadrant

Solution: Given, \sec x = \frac{13}{5}

Sinc, \cos x =\frac{1}{\sec x}

\Rightarrow \cos x = \frac{1}{13/5}=\frac{5}{13}

We know that

\sin^2 x = 1-\cos^2 x

\Rightarrow \sin^2x = 1-(\frac{5}{13})^2

= 1-\frac{25}{169}=\frac{144}{169}

\Rightarrow \sin x= \pm\frac{12}{13}

x lies in fourth quadrant

\sin x = -\frac{12}{13}

\operatorname{cosec x}=\frac{1}{\sin x}=-\frac{13}{12}

Since, \tan x =\frac{\sin x}{\cos x} = \frac{-12/13}{5/13}

\Rightarrow \tan x = -\frac{12}{5}

Since, \cot x = \frac{1}{\tan x} = -\frac{5}{12}

Question 5: \tan x = -\frac{5}{12}, x lies in second quadrant.

Solution: Given \tan x = -\frac{5}{12}

Since \cot x =\frac{1}{\tan x} = -\frac{12}{5}

We know that

\sec^2x = 1+\tan^2x

\Rightarrow \sec^2 x = 1+(-\frac{5}{12})^2

= 1 +\frac{25}{144} =\frac{169}{144}

\sec x = \pm \frac{13}{12}

x lies in second quadrant

\sec x = -\frac{13}{12}

Now, \cos x = \frac{1}{\sec x}=-\frac{12}{13}

Again we know that

\sin^2x = 1-\cos^2x

\Rightarrow \sin^2x = 1-(-\frac{12}{13})^2

= 1-\frac{144}{169}= \frac{25}{169}

\Rightarrow \sin x=\pm\frac{5}{13}

x lies in second quadrant

\therefore \sin x = \frac{5}{13}

\operatorname{cosec x} = \frac{1}{\sin x} = \frac{13}{5}

Find the values of the trigonometric functions in Exercises 6 to 10.

Question 6: sin 765°

Solution: The values of sin x repeat after an interval of 2π or 360°

So we get

\sin 765^{\circ}=\sin(45+2\times 360)

= \sin 45^{\circ}

=\frac{1}{\sqrt{2}}

question 7: \operatorname{cosec}(-1410)

Solution: The values of sin x repeat after an interval of 2π or 360°

\operatorname{cosec}(-1410)=\operatorname{cosec}(-1410+4\times 360)

=\operatorname{cosec}(30)

=2

Question 8: \tan(\frac{19\pi}{3})

Solution: The values of sin x repeat after an interval of 2π or 360°

\tan(\frac{19\pi}{3})=\tan(\frac{19\pi}{3}-3\times 2\pi)

=\tan(\frac{19\pi}{3}-6\pi)

= \tan(\frac{\pi}{3})

= \frac{1}{\sqrt{3}}

Question 9: \sin(-\frac{11\pi}{3})

Solution: The values of sin x repeat after an interval of 2π or 360°

\sin(-\frac{11\pi}{3})=\sin(-\frac{11\pi}{3}+2\times 2\pi)

= \sin(\frac{\pi}{3})

= \frac{\sqrt{3}}{2}

Question 10: \cot(-\frac{15\pi}{4})

Solution: The values of sin x repeat after an interval of 2π or 360°

\cot(-\frac{15\pi}{4})=\cot(-\frac{15\pi}{4}+2\times 2\pi)

=\cot(-\frac{15\pi}{4}+4\pi)

=\cot(\frac{\pi}{4})

=1


 

https://gmath.in/ex-3-1-trigonometric-function-ncert-solution-class-11/

 

class 12 inverse trigonometric functions multiple choice

 

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