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NCERT Solutions for Miscellaneous Exercise Chapter 2 Class 12 - Inverse Trigonometric Functions

NCERT Solutions for Miscellaneous Exercise Chapter 2 Class 12 - Inverse Trigonometric Functions

Edited By Ramraj Saini | Updated on Dec 04, 2023 12:06 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 2 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 2 class 12 Inverse Trigonometric Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise has a set of questions which are not introduced in earlier exercises. Class 12 Maths chapter 2 miscellaneous exercise basically deals with the questions related to proving of the inverse trigonometric functions equivalent. By using various methods and techniques, such equations are proved equal to each other. Students can easily find that many questions were asked from Class 12 Maths chapter 2 miscellaneous exercise in the board exams. Hence it is highly recommended to practice the NCERT solutions for Class 12 maths chapter 2 including miscellaneous exercise which is present in NCERT Class 12th book.

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Miscellaneous exercise class 12 chapter 2 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Miscellaneous Exercise

Question:1 Find the value of the following: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

Answer:

If x \epsilon [0,\pi] then \cos^{-1}(\cos x) = x , which is principal value of \cos^{-1} x .

So, we have \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right )

where \:\frac{13\pi}{6} \notin \left [ 0, \pi \right ].

Hence\: we \:can\: write\: \cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) \:as

=\cos^{-1}\left (\cos\left(2\pi+\frac{\pi}{6} \right ) \right )

=\cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right )

\frac{\pi}{6}\ \epsilon \left [ 0, \pi \right ]

Therefore we have,

\cos^{-1}\left (\cos\left(\frac{13\pi}{6} \right ) \right ) = \cos^{-1}\left (\cos\left(\frac{\pi}{6} \right ) \right ) = \frac{\pi}{6} .

Question:2 Find the value of the following: \tan^{-1}\left(\tan\frac{7\pi}{6} \right )

Answer:

We have given \tan^{-1}\left(\tan\frac{7\pi}{6} \right ) ;

so, as we know \tan^{-1}\left(\tan x \right ) =x \:\:if\:\:x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

So, here we have \frac{7\pi}{6} \notin \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ) .

Therefore we can write \tan^{-1}\left(\tan\frac{7\pi}{6} \right ) as:

=\tan^{-1}\left(\tan \left (2\pi - \frac{5\pi}{6} \right ) \right ) \left [ \because \tan(2\pi - x) = -\tan x \right ]

=\tan^{-1}\left[-\tan \left ( \frac{5\pi}{6} \right ) \right ]

=\tan^{-1}\left[\tan \left (\pi- \frac{5\pi}{6} \right ) \right ]

=\tan^{-1}\left[\tan \left (\frac{\pi}{6} \right ) \right ]\:\:where\:\:\frac{\pi}{6} \epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

\therefore \tan^{-1} \left ( \tan \frac{7\pi}{6} \right ) = \tan^{-1} \left ( \tan \frac{\pi}{6} \right ) = \frac{\pi}{6} .

Question:3 Prove that 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}

Answer:

To prove: 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7} ;

L.H.S=2\sin^{-1}\frac{3}{5}

Assume that \sin^{-1}\frac{3}{5} = x

then we have \sin x = \frac{3}{5} .

or \cos x = \sqrt{1-\left (\frac{3}{5} \right )^2} = \frac{4}{5}

Therefore we have

\tan x = \frac{3}{4}\:\:or\:\:x = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}

Now,

We can write L.H.S as

2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4}

=\tan^{-1} \left [\frac{2\times\frac{3}{4}}{1- \left ( \frac{3}{4} \right )^2} \right ] as we know \left [2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \right ]

=\tan^{-1} \left [\frac{\frac{3}{2}}{\left ( \frac{16-9}{16} \right )} \right ] = \tan^{-1}\left ( \frac{3}{2}\times \frac{16}{7} \right )

=\tan^{-1} \frac{24}{7}=R.H.S

L.H.S = R.H.S

Question:4 Prove that \sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} =\tan^{-1}\frac{77}{36}

Answer

Taking \sin ^{-1} \frac{8}{17} = x

then,

\sin x = \frac{8}{17} \Rightarrow \cos x = \sqrt{1- \left ( \frac{8}{17} \right )^2} = \sqrt {\frac{225}{289}} = \frac{15}{17}.

Therefore we have-

\tan^{-1} x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15}

\therefore \sin ^{-1} \frac{8}{17} = \tan ^{-1} \frac{8}{15} .............(1).

Now, let\:\sin ^{-1} \frac{3}{5} = y ,

Then,

\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4} .............(2).

So, we have now,

L.H.S.

\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5}

using equations (1) and (2) we get,

=\tan ^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}

=\tan^{-1} \frac{\frac{8}{15}+ \frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}} [\because \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1 - xy}]

=\tan^{-1} (\frac{32+45}{60-24})

=\tan^{-1} (\frac{77}{36})

= R.H.S.

Question:5 Prove that \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}

Answer:

Take \cos^{-1}\frac{4}{5} = x and \cos^{-1}\frac{12}{13} = y and \cos^{-1}\frac{33}{65} = z

then we have,

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

\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}

Then we can write it as:

\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} or x= \tan^{-1} \frac{3}{4}

\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4} ...............(1)

Now, \cos^{-1}\frac{12}{13} = y

\cos y = \frac{12}{13} \Rightarrow \sin y =\frac{5}{13}

\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}

So, \cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12} ...................(2)

Also we have similarly;

\cos^{-1}\frac{33}{65} = z

Then,

\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33} ...........................(3)

Now, we have

L.H.S

\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} so, using (1) and (2) we get,

=\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}

=\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right ) \because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]

=\tan^{-1}\left ( \frac{36+20}{48-15} \right )

=\tan^{-1}\left ( \frac{56}{33} \right ) or we can write it as;

=\cos^{-1}\frac{33}{65}

= R.H.S.

Hence proved.


Question:6 Prove that \cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}

Answer:

Converting all terms in tan form;

Let \cos^{-1}\frac{12}{13} = x , \sin^{-1}\frac{3}{5} = y and \sin^{-1}\frac{56}{65} = z .

now, converting all the terms:

\cos^{-1}\frac{12}{13} = x or \cos x = \frac{12}{13}

We can write it in tan form as:

\cos x = \frac{12}{13} \Rightarrow \sin x = \frac{5}{13} .

\therefore \tan x = \frac{5}{12} \Rightarrow x = \tan^{-1} \frac{5}{12}

or \cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12} ................(1)

\sin^{-1}\frac{3}{5} = y or \sin y = \frac{3}{5}

We can write it in tan form as:

\sin y = \frac{3}{5} \Rightarrow \cos y = \frac{4}{5}

\therefore \tan y =\frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4}

or \sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4} ......................(2)

Similarly, for \sin^{-1}\frac{56}{65} = z ;

we have \sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33} .............(3)

Using (1) and (2) we have L.H.S

\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}

= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{3}{4}

On applying \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}

We have,

=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}

=\tan^{-1} (\frac{20+36}{48-15})

=\tan^{-1} (\frac{56}{33})

=\sin^{-1} (\frac{56}{65}) ...........[Using (3)]

=R.H.S.

Hence proved.

Question:7 Prove that \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}

Answer:

Taking R.H.S;

We have \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}

Converting sin and cos terms in tan forms:

Let \sin^{-1}\frac{5}{13} = x and \cos^{-1}\frac{3}{5} = y

now, we have \sin^{-1}\frac{5}{13} = x or \sin x = \frac{5}{13}

\sin x = \frac{5}{13} \:or\: \cos x =\frac{12}{13}\:or\:\tan x = \frac{5}{12}

\tan x = \frac{5}{12} \Rightarrow x =\tan^{-1} \frac{5}{12}

\therefore \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12} ............(1)

Now, \cos^{-1}\frac{3}{5} = y\Rightarrow \cos y = \frac{3}{5}

\cos y = \frac{3}{5} \:or\: \sin y = \frac{4}{5}\:or\:\tan y = \frac{4}{3}

\tan y = \frac{4}{3} \Rightarrow y = \tan^{-1} \frac{4}{3}

\therefore \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{4}{5} ................(2)

Now, Using (1) and (2) we get,

R.H.S.

\sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}

=\tan^{-1}\left ( \frac{\frac{5}{12}+\frac{4}{3}}{1- \frac{5}{12}\times \frac{4}{3}} \right ) as we know \left [ \tan^{-1} x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} \right ]

so,

= \tan^{-1} \frac{63}{16}

equal to L.H.S

Hence proved.

Question:8 Prove that \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} +\tan^{-1}\frac{1}{3} +\tan^{-1}\frac{1}{8} = \frac{\pi}{4}

Answer:

Applying the formlua:

\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy} on two parts.

we will have,

=\tan^{-1}\left (\frac{\frac{1}{5}+ \frac{1}{7}}{1- \frac{1}{5}\times \frac{1}{7}} \right ) + \tan^{-1}\left (\frac{\frac{1}{3}+ \frac{1}{8}}{1- \frac{1}{3}\times \frac{1}{8}} \right )

= \tan^{-1} \left ( \frac{7+5}{35-1} \right ) + \tan^{-1} \left ( \frac{8+3}{24-1} \right )

= \tan^{-1} \left ( \frac{12}{34} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )

= \tan^{-1} \left ( \frac{6}{17} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )

= \tan^{-1}\left [ \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times\frac{11}{23}} \right ]

= \tan^{-1}\left [ \frac{325}{325} \right ] = \tan^{-1} 1

=\frac{\pi}{4}

Hence it s equal to R.H.S

Proved.

Question:9 Prove that \tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]

Answer:

By observing the square root we will first put

x= \tan^2 \theta .

Then,

we have \tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}

or, R.H.S.

\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)

= \frac{1}{2}\times 2\theta = \theta .

L.H.S. \tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta

hence L.H.S. = R.H.S proved.

Question:10 Prove that \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \frac{x}{2},\;\;x\in\left(0,\frac{\pi}{4} \right )

Answer:

Given that \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )

By observing we can rationalize the fraction

\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right )

We get then,

=\left(\frac{\sqrt{1+\sin x} + \sqrt{1 - \sin x}}{\sqrt{1+\sin x} - \sqrt{1 - \sin x}} \right ) = \left(\frac{(\sqrt{1+\sin x} + \sqrt{1 - \sin x})^2}{{1+\sin x} - {1 + \sin x}} \right )

= \left(\frac{{1+\sin x} +{1 - \sin x} + 2\sqrt{(1+\sin x)(1-\sin x)} }{{1+\sin x} - {1 + \sin x}} \right )

= \frac{2(1+\sqrt{1-\sin^2 x})}{2\sin x} = \frac{1+\cos x}{\sin x} = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}

= \cot \frac{x}{2}

Therefore we can write it as;

\cot^{-1}\left ( \cot \frac{x}{2} \right ) = \frac{x}{2}

As L.H.S. = R.H.S.

Hence proved.

Question:11 Prove that \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1

[Hint: Put x = \cos 2\theta ]

Answer:

By using the Hint we will put x = \cos 2\theta ;

we get then,

=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )

=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )

=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )

=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right ) dividing numerator and denominator by \cos \theta ,

we get,

= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )

= \tan^{-1} 1 - \tan^{-1} (\tan \theta) using the formula \left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]

= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x

As L.H.S = R.H.S

Hence proved

Question:12 Prove that \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}

Answer:

We have to solve the given equation:

\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}

Take \frac{9}{4} as common in L.H.S,

=\frac{9}{4}\left [ \frac{\pi}{2}- \sin^{-1}\frac{1}{3} \right ]

or =\frac{9}{4}\left [ \cos^{-1}\frac{1}{3} \right ] from \left [ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \right ]

Now, assume,

\left [ \cos^{-1}\frac{1}{3} \right ] = y

Then,

\cos y = \frac{1}{3} \Rightarrow \sin y = \sqrt{1-(\frac{1}{3})^2} = \frac{2.\sqrt2}{3}

Therefore we have now,

y = \sin^{-1} \frac{2.\sqrt2}{3}

So we have L.H.S then = \frac{9}{4}\sin^{-1} \frac{2.\sqrt2}{3}

That is equal to R.H.S.

Hence proved.

Question:13 Solve the following equations: 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x)

Answer:

Given equation 2\tan^{-1}(\cos x) = \tan^{-1}(2\textup{cosec}x) ;

Using the formula:

\left [ 2\tan^{-1}z = \tan^{-1} \frac{2z}{1-z^2} \right ]

We can write

2\tan^{-1}(\cos x) = \tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ]

\tan^{-1}\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \tan^{-1}\left [2cosec x \right ]

So, we can equate;

=\left [ \frac{2\cos x}{1- (\cos x )^2 }\right ] = \left [2cosec x \right ]

=\left [ \frac{2\cos x}{\sin^2 x }\right ] = \left [ \frac{2}{sinx } \right ]

that implies that \cos x = \sin x .

or \tan x =1 or x = \frac{\pi}{4}

Hence we have solution x = \frac{\pi}{4} .

Question:14 Solve the following equations: \tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x,\;(x>0)

Answer:

Given equation is

\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x :

L.H.S can be written as;

\tan^{-1} \frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x

Using the formula \left [ \tan^{-1}x -\tan^{-1}y = \tan^{-1} \frac{x-y}{1+xy} \right ]

So, we have \tan^{-1}1 - \tan^{-1}x = \frac{1}{2} \tan^{-1}x

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

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

\Rightarrow \tan^{-1}x = \frac{\pi}{6}

\Rightarrow x= \tan \frac{\pi}{6} = \frac{1}{\sqrt3}

Hence the value of x= \frac{1}{\sqrt3} .

Question:15 \sin(\tan^{-1}x),\;|x|<1 is equal to

(A) \frac{x}{\sqrt{1-x^2}}

(B) \frac{1}{\sqrt{1-x^2}}

(C) \frac{1}{\sqrt{1+x^2}}

(D) \frac{x}{\sqrt{1+x^2}}

Answer:

Let \tan^{-1}x = y then we have;

\tan y = x or

y=\sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right ) \Rightarrow \tan^{-1} x = \sin^{-1} \left ( \frac{x}{\sqrt{1+x^2}} \right)

\Rightarrow \sin \left ( \tan^{-1} x \right ) = \sin\left ( \sin^{-1}\left ( \frac{x}{\sqrt{1+x^2}} \right ) \right ) = \frac{x}{\sqrt {1+x^2}}

Hence the correct answer is D.

Question:16 \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2} then x is equal to

(A) 0,\frac{1}{2}

(B) 1,\frac{1}{2}

(C) 0

(D) \frac{1}{2}

Answer:

Given the equation: \sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}

we can migrate the \sin^{-1}(1-x) term to the R.H.S.

then we have;

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

or - 2\sin^{-1}x =\cos^{-1}(1-x) ............................(1)

from \left [\because \cos^{-1}(1-x) + \sin^{-1}(1-x) = \frac{\pi}{2} \right ]

Take \sin^{-1}x = \Theta \Rightarrow \sin \Theta = x or \cos \Theta = \sqrt{1-x^2} .

So, we conclude that;

\sin^{-1}x = \cos^{-1}\left ( \sqrt{1-x^2} \right )

Therefore we can put the value of \sin^{-1}x in equation (1) we get,

- 2\cos^{-1}\left ( \sqrt{1-x^2} \right ) =\cos^{-1}(1-x)

Putting x= sin y , in the above equation; we have then,

\Rightarrow - 2\cos^{-1}\left ( \sqrt{1-(\sin y)^2} \right ) =\cos^{-1}(1-\sin y )

\Rightarrow - 2\cos^{-1}\left ( \sqrt{\cos^2 y} \right ) =\cos^{-1}(1-\sin y )

\Rightarrow - 2\cos^{-1}\left ( \cos y \right ) =\cos^{-1}(1-\sin y )

\Rightarrow \cos(-2y) = 1-\sin y

\Rightarrow - 2y=\cos^{-1}(1-\sin y )

\Rightarrow 1- 2\sin^2 y = 1-\sin y

\Rightarrow 2\sin^2 y - \sin y = 0

\Rightarrow \sin y(2 \sin y -1) = 0

So, we have the solution;

\sin y = 0\ or\ \frac{1}{2} Therefore we have x = 0\ or\ x= \frac{1}{2} .

When we have x= \frac{1}{2} , we can see that :

L.H.S. = \sin ^{-1}\left ( 1 - \frac{1}{2} \right ) - 2\sin^{-1}\frac{1}{2} = - \sin^{-1}\frac{1}{2} = -\frac{\pi}{6}

So, it is not equal to the R.H.S. -\frac{\pi}{6} \neq \frac{\pi}{2}

Thus we have only one solution which is x = 0

Hence the correct answer is (C).

Question:17 \tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1}\frac{x-y}{x+y} is equal to

(A) \frac{\pi}{2}

(B) \frac{\pi}{3}

(C) \frac{\pi}{4}

(D) \frac{3\pi}{4}

Answer:

Applying formula: \left [ \tan^{-1} x - \tan^{-1}y = \tan^{-1} \left ( \frac{x-y}{1+xy} \right ) \right ] .

We get,

\tan^{-1}\left (\frac{x}{y} \right )-\tan^{-1} \left ( \frac{x-y}{x+y} \right ) = \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ]

= \tan^{-1} \left [\frac{ \frac{x}{y} - \frac{x-y}{x+y}}{1+\left ( \frac{x}{y} \right ) \left ( \frac{x-y}{x+y} \right ) } \right ] = \tan^{-1} \left [ \frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y) + x(x-y)}{y(x+y)}} \right ]

= \tan^{-1}\left ( \frac{x^2+xy - xy + y^2}{xy + y^2 + x^2 - xy} \right )

= \tan^{-1}\left ( \frac{x^2 + y^2}{ y^2 + x^2 } \right ) = \tan^{-1} 1 = \frac{\pi}{4}

Hence, the correct answer is C.

More About NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise

The NCERT Class 12 Maths chapter Inverse Trigonometric functions provided here is prepared by the experienced faculties. NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise covers the major syllabus of this chapter from exam perspective. As questions from this exercise are asked more than those of previous exercises. Therefore NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercise becomes a must to do exercise for the examination.

Also Read| Inverse Trigonometric Functions NCERT Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercises

  • Miscellaneous exercise chapter 2 Class 12 NCERT syllabus has some of the questions which are very important from exam point of view.
  • Questions mentioned in NCERT book Class 12 Maths chapter 2 miscellaneous solutions are of the level of JEE and NEET.
  • It is a good source for revision also. Hence NCERT solutions for Class 12 Maths chapter 2 miscellaneous exercises can be referred directly to score well in the exam.
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Key Features Of NCERT Solutions For Class 12 Chapter 2 Miscellaneous Exercise

  • Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 2, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 chapter 2 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this class 12 maths ch 2 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for class 12 chapter 2 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject wise NCERT Exemplar solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What is the importance of miscellaneous exercise from exam perspective?

As direct questions are asked in the exam from this exercise, it is important to practice miscellaneous exercise before the examination. For more questions students can use NCERT exemplar.

2. List out the important topics of NCERT text book Class 12 Maths chapter 2?

The topics which are Important are among the following 

  • finding the inverse of sine, cos, tan etc. are important which are asked frequently in the exam

3. What is the overall importance of NCERT exercise in board examination ?

NCERT exercises are the favorite source of the Board examination. Hence it is advisable to go through the NCERT exercise. 

4. Does memorization help in solving proof related questions?

Basic values of inverse trigonometric functions can be memorized, rest you will have to brainstorm in proof related questions.  

5. What can be a basic approach to solve proof related questions ?

Process in step by step manner keeping in mind the final question can help in proving the desired direction. 

6. How many total exercises are there in Class 12 Maths chapter 2 ?

In NCERT Class 12 Maths chapter 2, there are a total of 3 exercises.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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October 23, 2024 - 04:01 PM IST :
CBSE Board Exam Date Sheet 2025 (OUT) Live: CBSE Classes 10, 12 time table out 86 days before exams start
November 20, 2024 - 11:17 PM IST
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November 20, 2024 - 09:56 PM IST
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