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NCERT Solutions for Exercise 3.2 Class 11 Maths Chapter 3 - Trigonometric Functions

NCERT Solutions for Exercise 3.2 Class 11 Maths Chapter 3 - Trigonometric Functions

Edited By Sumit Saini | Updated on Jul 29, 2022 03:15 PM IST

NCERT solutions for class 11 maths chapter 3 trigonometric functions exercise 3.2 deal with finding the values of trigonometric functions at different quadrants. Also exercise 3.2 class 11 maths deal with finding values of other trigonometric functions if one function value is given. This is the very basic exercise of class 11 maths NCERT text book. The NCERT syllabus is helpful for exams like JEE Main. Other than NCERT class 11 maths exercise 3.2 following exercise are also present

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.2

Question:1 Find the values of other five trigonometric functions \small \cos x = -\frac{1}{2} , x lies in third quadrant.

Answer:

Solution
\cos x = -\frac {1}{2}
\because \sec x = \frac{1}{\cos x} = \frac{1}{-\frac {1}{2}} = -2
x lies in III quadrants. Therefore sec x is negative

\sin ^{2}x +\cos^{2}x = 1 \\ \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 = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}
x lies in III quadrants. Therefore sin x is negative
\therefore \sin x= - \frac{\sqrt{3}}{2}

\because cosec \ x = \frac {1}{\sin x}= \frac{1}{- \frac{\sqrt{3}}{2}} =- \frac{2}{\sqrt{3}}

x lies in III quadrants. Therefore cosec x is negative

\tan x = \frac{\sin x}{\cos x} = \frac {-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}
x lies in III quadrants. Therefore tan x is positive

\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}
x lies in III quadrants. Therefore cot x is positive

Question:2 Find the values of other five trigonometric functions \small \sin x = \frac{3}{5} x lies in second quadrant.

Answer:

Solution

\sin x = \frac {3}{5}

cosec \ x = \frac{1}{\sin x}=\frac {1}{\frac {3}{5}} = \frac {5}{3}
x lies in the second quadrant. Therefore cosec x is positive

\sin^{2}x + \cos ^{2}x = 1\\ \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 = \sqrt{\frac {16}{25}} = \pm \frac {4}{5}
x lies in the second quadrant. Therefore cos x is negative
\therefore \cos x = - \frac {4}{5}

\sec x = \frac {1}{\cos x} = \frac{1}{- \frac {4}{5}} = -\frac {5}{4}
x lies in the second quadrant. Therefore sec x is negative

\tan x = \frac {\sin x}{\cos x} = \frac {\frac{3}{5}}{-\frac{4}{5}} = -\frac {3}{4}
x lies in the second quadrant. Therefore tan x is negative

\cot x = \frac {1}{\tan x} = \frac {1}{-\frac {3}{4}} = -\frac{4}{3}
x lies in the second quadrant. Therefore cot x is negative

Question:3 Find the values of other five trigonometric functions \small \cot x = \frac{3}{4} , x lies in third quadrant.

Answer:

Solution

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

\tan x = \frac{1}{\cot x}= \frac{1}{\frac {3}{4}} = \frac {4}{3}
1 + \tan ^ {2}x = \sec ^{2}x\\ 1+\frac{4^2}{3^2} = \sec ^{2}x\\ \\ 1 + \frac {16}{9} = \sec ^{2}x\\ \frac {25}{9} = \sec ^{2}x\\ \sec x = \sqrt {\frac {25}{9}} = \pm \frac {5}{3}
x lies in x lies in third quadrant. therefore sec x is negative
\sec x = -\frac{5}{3}

\cos x = \frac {1}{\sec x} = \frac {1}{-\frac{5}{3}} = -\frac{3}{5}
\sin ^{2 }x+ \cos ^{2}x = 1\\ \sin ^{2 }x = 1 - \cos ^{2}x\\ \sin ^{2 }x = 1 -\left ( -\frac{3}{5} \right )^{2}\\ \sin ^{2 }x = 1 - \frac {9}{25}\\ \sin ^{2 }x = \frac{16}{25}\\ \sin x = \sqrt {\frac{16}{25}} = \pm \frac{4}{5}
x lies in x lies in third quadrant. Therefore sin x is negative
\sin x = -\frac {4}{5}
cosec x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}

Question:4 Find the values of other five trigonometric functions \small \sec x = \frac{13}{5} , x lies in fourth quadrant.

Answer:

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\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \frac {5}{13}\\ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}\\ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}
lies in fourth quadrant. Therefore sin x is negative
\sin x =- \frac {12}{13}
\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12}
\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}
\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}

Question:5 Find the values of the other five trigonometric functions \small \tan x = -\frac{5}{12} , x lies in second quadrant.

Answer:
\tan x = -\frac {5}{12}
\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}
1 + \tan^{2}x = \sec^{2}x\\ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x\\ 1 + \frac {25}{144} = \sec^{2}x\\ \\ \frac {169}{144} = \sec^{2}x\\ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}
x lies in second quadrant. Therefore the value of sec x is negative
\sec x = - \frac {13}{12}
\cos x = \frac{1}{\sec x}= \frac{1}{-\frac{13}{12}} = -\frac {12}{13}
\sin^{2}x + \cos^{2}x = 1\\ \sin^{2}x = 1 - \cos^{2}x\\ \sin^{2}x = 1 - \left ( -\frac{12}{13} \right )^{2}\\ \sin^{2}x = 1 - \frac{144}{169}\\ \sin^{2}x = \frac {25}{169}\\ \sin x = \sqrt {\frac{25}{169}} = \pm \frac{5}{13}
x lies in the second quadrant. Therefore the value of sin x is positive
\sin x = \frac {5}{13}
\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}

Question:6 Find the values of the trigonometric functions \small \sin 765\degree

Answer:
We know that values of sin x repeat after an interval of 2\pi\ or\ 360\ degree

\sin765\degree = \sin (2\times360\degree + 45\degree ) = \sin45\degree\\ sin45\degree = \frac {1}{\sqrt{2}}

Question:7 Find the values of the trigonometric functions \small cosec \ (-1410\degree)

Answer:

We know that value of cosec x repeats after an interval of 2\pi \ or \ 360\degree
cosec (-1410\degree) = cosec (-1410\degree + 360\degree\times4)\\ cosec\ 30\degree = 2

or

cosec(-1410\degree)= - cosec(1410\degree)\\= -cosec(4 \times 360\degree - 30\degree)= - cosec(-30\degree) = 2

Question:8 Find the values of the trigonometric functions \small \tan \frac{19\pi }{3}

Answer:

We know that tan x repeats after an interval of \pi or 180 degree
\tan (\frac{19\pi}{3}) = \tan (6\pi+\frac{\pi}{3})= \tan \frac{\pi}{3} =\tan 60\degree = \sqrt{3}

Question:9 Find the values of the trigonometric functions \sin\left ( -\frac{11\pi}{3} \right )

Answer:

We know that sin x repeats after an interval of 2\pi or 360\degree
\sin \left ( -\frac{11\pi}{3} \right ) = \sin \left ( -4\pi +\frac{\pi}{3} \right ) = \sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}

Question:10 Find the values of the trigonometric functions \small \cot \left ( -\frac{15\pi }{4} \right )

Answer:

We know that cot x repeats after an interval of \pi or 180\degree
\cot \left ( -\frac{15\pi}{4} \right ) = \cot \left ( -4\pi +\frac {\pi}{4} \right ) = \cot \left ( \frac{\pi}{4} \right ) = 1


Also see-

NCERT solutions of class 11 subject wise

Subject wise NCERT Exampler solutions

Frequently Asked Question (FAQs)

1. Mention the concept of trigonometry present Exercise 3.2 Class 11 Maths?

Questions based on finding the trigonometric functions when a function is already provided.

2. Apart from maths, in which subject trigonometry is used ?

In Physics, trigonometry is used in some of the chapters.

3. How many questions can one expect from trigonometry in the CBSE exam ?

2 questions for sure will be there in the final exams. More can be there.

4. Can one skip this chapter as it is difficult to understand ?

No, this chapter has linkages in other chapters also. So it cannot be skipped.

5. What is the number of multiple choice questions provided in this exercise ?

Unlike previous exercises, this exercise does not have any questions based on multiple choice.

6. Are difficult questions asked from this chapter in the CBSE exam ?

Some questions can be difficult to answer but most of them will be easy.

7. How much time should one devote to complete Exercise 3.2 class 11 maths ?

For the first time around 2 to 3 hours are sufficient to complete this exercise.

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