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In advanced mathematics, inverse trigonometric functions play a major role, mainly in calculus and coordinate geometry. In the NCERT class 12 Maths chapter 2, the miscellaneous exercise combines different concepts from the chapter to help the students get an overall competency of the whole chapter. The students will be able to enhance their understanding of the chapter and get better at problem-solving. In this article of the NCERT Solutions for the Miscellaneous exercise of chapter 2 in class 12 maths, we will provide clear and step-by-step solutions for the exercise problems and help the students build their confidence in mathematics, so that they can prepare for various examinations. The latest guidelines of NCERT have been followed in this article.
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
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
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
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.
Also Read,
The main topics covered in Chapter 2 of inverse trigonometric functions, miscellaneous exercises are:
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$
Also, read,
Given below are some useful links for subject-wise NCERT solutions of class 12.
Here are some links to subject-wise solutions for the NCERT exemplar class 12.
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.
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
NCERT exercises are the favorite source of the Board examination. Hence it is advisable to go through the NCERT exercise.
Basic values of inverse trigonometric functions can be memorized, rest you will have to brainstorm in proof related questions.
Process in step by step manner keeping in mind the final question can help in proving the desired direction.
In NCERT Class 12 Maths chapter 2, there are a total of 3 exercises.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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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!
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:
Re-evaluate Your Study Strategies:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
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.
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!
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.
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