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RD Sharma Class 12 Exercise 3.8 Inverse Trigonometry functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.8 Inverse Trigonometry functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:33 PM IST

The CBSE Board students widely use the RD Sharma Solution books to clear their doubts in their homework and score good marks in their public examinations. However, every student cannot depend on a tutor to help them clear their doubts 24 x 7. Significantly, the students who find it difficult to cope with the Inverse Trigonometry portions can use the RD Sharma Class 12th exercise 3.8 books.

RD Sharma Class 12 Solutions Chapter3 Inverse Trigonometric Functions-Other Exercise

Inverse Trigonometric Functions Excercise:3.8

Inverse trigromtery functions exercise 3.8 question 1 (I)

Hint:
As we know that the value of \sin \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have,
\sin \left(\sin ^{-1} \frac{7}{25}\right)
Solution:
So here in the expression already present the form of \sin \left(\sin ^{-1} x\right)
\sin \left(\sin ^{-1} \frac{7}{25}\right)=\frac{7}{25} \mid

Inverse trigromtery functions exercise 3.8 question 1 (II)

Answer:

\frac{12}{13}
Hint:
As we know that the value of \sin \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have,
\sin \left(\cos ^{-1} \frac{5}{13}\right)
Solution:
Here in the place of \sin ^{-1} x there is \cos ^{-1} x.
So we convert \cos ^{-1} x into \sin ^{-1} x.
Let’s suppose that
\begin{aligned} &\cos ^{-1} \frac{5}{13}=\alpha \\ &\cos \alpha=\frac{5}{13} \\ &\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha \end{aligned}
Let, in\Delta ABC, angle CAB = \alpha and right angle at B
\cos \alpha=\frac{5}{13}=\frac{B}{H}

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=5^{2}+(B C)^{2} \\ &(B C)^{2}=169-25 \end{aligned}
BC= \pm 12 [we will ignore the -ve sign because BC is a length and it can’t be -ve.]
BC= 12
\sin =\frac{12}{13}
So, we required \alpha and that is \frac{12}{13}
\sin \left(\cos ^{-1} \frac{5}{13}\right)=\sin \alpha=\frac{12}{13}

Inverse trigromtery functions exercise 3.8 question 1 (III)

Answer:

\frac{24}{25}
Hint:
As we know that the value of \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have \sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)
Solution:
Here in the place of \sin ^{-1} xwe have \tan ^{-1} x.
So, we convert \tan ^{-1} x to \sin ^{-1} x.
Let’s suppose that,
\begin{aligned} &\tan ^{-1}\left(\frac{24}{7}\right)=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} \tan \alpha &=\frac{P}{B}=\frac{24}{7} \\ (A C)^{2} &=(A B)^{2}+(B C)^{2} \\ &=7^{2}+(24)^{2} \\ &=49+576 \\ &=625 \end{aligned}
= +25 [we will ignore the -ve sign because AC is a length and it can’t be -ve.]
AC= 25
\begin{aligned} &\sin \alpha=\frac{24}{25} \\ &\sin \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\sin \alpha=\frac{24}{25} \end{aligned} [since, \tan ^{-1}\left(\frac{24}{7}\right)=\alpha]

Inverse trigonometric functions exercise 3.8 question 1 (iv)

Answer:

\frac{15}{17}

Hint:
As we know that the value of \left(\sin ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have
\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)
Solution:
So here in place of \sin ^{-1} x there is \sec ^{-1} x.
So, we convert \sec ^{-1} x into \sin ^{-1} x
Let’s we suppose that,
\begin{aligned} &\sec ^{-1} \frac{17}{8}=\alpha \\ &\sec \alpha=\frac{17}{8} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &\sec \alpha=\frac{H}{B}=\frac{17}{8} \\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \end{aligned}
BC =\pm 15 [we will ignore the -ve sign because BC is a length and it can’t be -ve]
BC = 15
\begin{aligned} &\sin \alpha=\frac{P}{H}=\frac{15}{17} \\ &\sin \left(\sec ^{-1}\left(\frac{17}{8}\right)\right)=\frac{15}{17} \end{aligned}

Inverse trigonometric functions exercise 3.8 question 1 (v)

Answer:

\frac{5}{4}
Hint:
As we know that the value of \cos e c\left(\cos e c^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)
Given:
We have
\operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)
Solution:
In place of \cos e c^{-1} x here is \cos ^{-1} x
So, we convert \cos ^{-1} x into \cos e c^{-1} x.
Let’s suppose that,
\begin{aligned} &\cos ^{-1} \frac{3}{5}=\alpha \\ &\cos \alpha=\frac{3}{5}=\frac{B}{H} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &5^{2}=3^{2}+(B C)^{2} \\ &25-9=(B C)^{2} \end{aligned}
BC =\pm 4 [we will ignore the -ve sign because BC is a length and it can’t be -ve]
BC =4
\begin{aligned} &\cos e c \alpha=\frac{H}{P}=\frac{5}{4} \\ &\cos e c\left(\cos ^{-1} \frac{3}{5}\right)=\frac{5}{4} \end{aligned}

Inverse trigonometric functions exercise 3.8 question 1 (vi)

Answer:

\frac{13}{5}

Hint:
As we know that the range of \sec \left(\sec ^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)
Given:
We have
\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)
Solution:
In place of \sec ^{-1} x there is \sin ^{-1} x
So, we convert \sin ^{-1} x into \sec ^{-1} x .
Let’s suppose that,

\begin{aligned} &\sin ^{-1} \frac{12}{13}=\alpha \\ &\sin \alpha=\frac{12}{13} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=(AB)^{2}+(12)^{2} \\ &(AB)^2=25 \end{aligned}
AB = \pm 5 [we will ignore the -ve sign because AB is a length and it can’t be -ve]
AB = 5
\begin{aligned} &\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\sec \alpha=\frac{13}{5} \\ &\sec \left(\sin ^{-1}\left(\frac{12}{13}\right)\right)=\frac{13}{5} \end{aligned}

Inverse Trigonometric functions exercise 3.8 question 1 (vii)

Answer:

\frac{15}{8}

Hint:
As we know that the value of \tan \left(\tan ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }
Given:
We have
\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)
Solution:
In place of \tan ^{-1} x here is \cos ^{-1} x
So, we convert \cos ^{-1} x into \tan ^{-1} x.
Let’s suppose that,
\begin{aligned} &\cos ^{-1} \left (\frac{8}{17} \right )=\alpha \\ &\cos \alpha=\frac{8}{17}=\frac{B}{H} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.


\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(17)^{2}=8^{2}+(B C)^{2} \\ &(B C)^{2}=289-64 \\ & (BC)^2= 25\end{aligned}
BC=\pm 15 [we will ignore the -ve sign because BC is a length and it can’t be -ve]
BC= 15
\begin{aligned} &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\tan \alpha \\ &\tan \alpha=\frac{15}{8} \\ &\tan \left(\cos ^{-1}\left(\frac{8}{17}\right)\right)=\frac{15}{8} \end{aligned}

Inverse Trigonometric functions exercise 3.8 question 1 (viii)

Answer:

\frac{3}{4}

Hint:
As we know that the range of \cot \left(\cot ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }
Given:
\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)
Solution:
In place of \cot ^{-1} x here is \cos ^{-1} x
So, we convert \cos ^{-1} x into \cot ^{-1} x.
Let’s suppose that,
\begin{aligned} &\cos ^{-1} \left (\frac{3}{5} \right )=\alpha \\ &\cos \alpha=\frac{3}{5}=\frac{B}{H} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=3^{2}+(B C)^{2} \\ &25=9 + (BC)^2 \\ & (BC)^2= 16\end{aligned}
BC=\pm 4 [we will ignore the -ve sign because BC is a length and it can’t be -ve]
BC= 4\begin{aligned} &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\cot c \\ &\cot \alpha=\frac{3}{4} \\ &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{4} \end{aligned}


Inverse Trigonometric functions exercise 3.8 question 1 (ix)

Answer:

\frac{7}{25}
Hint:
As we know that the range of \left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have
\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)
Solution:
In place of \cos ^{-1} x here is \tan ^{-1} x
So, we convert \tan ^{-1} x into \cos ^{-1} x .
Let’s suppose that,
\begin{aligned} &\tan ^{-1} \left (\frac{24}{7} \right )=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(AC)^{2}=7^{2}+(24)^{2} \\ &(A C)^{2}=625 \end{aligned}
AC=\pm 25 [we will ignore the -ve sign because AB is a length and it can’t be -ve]
AC=25
\begin{aligned} &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\cos \alpha \\ &\cos \alpha=\frac{7}{25} \\ &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\frac{7}{25} \end{aligned}

Inverse Trigonometric functions exercise 3.8 question 2 (i)

Hint:
We convert in the form of \tan (\alpha + \beta) so we know the formula of it.
Given:
We have to prove that \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6}
Solution:
LHS= \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)
Let suppose that \cos ^{-1}\frac{4}{5}=\alpha and \tan ^{-1}\frac{2}{3}=\beta,
\begin{aligned} &\cos \alpha=\frac{4}{5} \text { and } \tan \beta=\frac{2}{3} \\ &\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \end{aligned}
But we have not \tan \alpha. We have only \tan \beta. So, we convert \cos \alpha into \tan \alpha.
\cos \alpha =\frac{4}{5}=\frac{B}{H}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.


\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=4^{2}+(B C)^{2} \\ &(B C)^{2}=25-16 \\ \end{aligned}
BC =\pm 9
BC = 3
\tan \alpha = \frac{3}{4}

[\because \tan \alpha = \frac{3}{4} \; and\: \tan \beta = \frac{2}{3} ]
\begin{gathered} \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \\ =\frac{\frac{9+8}{12-6}}{12}=\frac{17}{6}=\mathrm{RHS} \\ \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6} \end{gathered} [ we take LCM of 3,4]
Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (ii)

Hint:
We convert in the form of \cos (\alpha + \beta)and we know the formula of it.
Given:
We have to prove \cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{6}{5 \sqrt{13}}
Solution:
LHS = \cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)
Let’s suppose that \sin ^{-1} \frac{3}{5}= \alphaand \cot ^{-1} \frac{3}{2}=\beta.
\sin \alpha = \frac{3}{5} \; and\; \cot \beta =\frac{3}{2}
\cos\left ( \alpha +\beta \right )= \cos \alpha \cos \beta -\sin \alpha \sin \beta
Here, we only have \sin \alpha and \cos \beta.So we have to find \cos \alphaand \sin \beta.
\sin \alpha =\frac{3}{5}=\frac{B}{H}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=(AB)^{2}+3^{2} \\ &(AB )^{2}=25-9 \\ \end{aligned}
AB=\pm 4
\cos \alpha = \frac{4}{5}
Let, in \Delta PQR, angle RPQ = \beta and right angle at B.
\cot \beta =\frac{3}{2}=\frac{B}{P}

\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(PR)^{2}=2^{2}+3^{2} \\ &(PR)^{2}=4 + 9 \\ \end{aligned}
PR =\pm \sqrt{13} [we will ignore the -ve sign because PR is a length and it can’t be -ve]
PR = \sqrt{13}
\begin{aligned} &P R=\sqrt{13} \\ &\sin \beta=\frac{2}{\sqrt{13}}, \cos \beta=\frac{3}{\sqrt{13}} \\ &\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}} \times \frac{3}{5} \\ &=\frac{12}{5 \sqrt{13}}-\frac{6}{5 \sqrt{13}} \\ &\cos (\alpha+\beta)=\frac{6}{5 \sqrt{13}}=\mathrm{RHS} \end{aligned}

Inverse trigonometric functions exercise 3.8 question 2 (iii)

Hint:
We convert expression in the form of \tan (\alpha + \beta) so we know the formula of it.
Given:
We have to prove \tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)=\frac{63}{16}
Solution:
LHS = \tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)
\sin ^{-1} \frac{5}{13}=\alpha \; and \; \cos ^{-1} \frac{3}{5}=\beta
\sin \alpha =\frac{5}{13} \; and \; \cos \beta = \frac{3}{5}
\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} …(i)
But we have not \tan \alpha and \tan \beta. So, we find out first then we put in equation (i)
\sin \alpha = \frac{5}{13}=\frac{P}H{}
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=AB^{2}+5^{2} \\ &(AB)^{2}=144 \\ \end{aligned}
AB =\pm 12 [we will ignore the -ve sign because AB is a length and it can’t be -ve]
AB =12
\therefore \tan \alpha = \frac{5}{12}

Let, in \Delta PQR, angle RPQ = \beta and right angle at B.
\cot \beta =\frac{3}{5}=\frac{B}{P}

\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(5)^{2}=3^{2}+(QR)^{2} \\ &(QR)^{2}=16\\ \end{aligned}
QR = \pm 4
QR = 4
\therefore \tan \beta = \frac{4}{3}
Now, \begin{aligned} \tan (\alpha+\beta)=& \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ &=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}} \end{aligned} [ put, \tan \alpha = \frac{5}{12} and \tan \beta = \frac{4}{3}]
[ we take LCM of 3 and 12]
\begin{aligned} &=\frac{\frac{5+16}{12}}{\frac{36-20}{36}} \\ &=\frac{\frac{21}{12}}{\frac{16}{36}} \\ &=\frac{21}{12} \times \frac{36}{16} \\ &=\frac{21}{1} \times \frac{3}{16} \\ \tan (\alpha+\beta) &=\frac{63}{16}=\mathrm{RHS} \end{aligned}
Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (iv)

Answer:

Hint:
We convert in the form of \sin (\alpha + \beta)and we know the formula of it.
Given:
We have to prove \sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\frac{63}{65}
Solution:
LHS = \sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)
Let’s suppose that \cos ^{-1} \frac{3}{5}=\alpha \; and \; \sin ^{-1} \frac{5}{13}=\beta
\cos \alpha= \frac{3}{5} \; and \; \sin \beta \frac{5}{13} and
\begin{aligned} &\Rightarrow \quad \sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right) \\ &\Rightarrow \quad \sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}
…(i)
Here we don’t have \sin \alpha . \cos \beta. So, we find out it and then put all the values in equation (i)
Let, in \Delta ABC, angle CAB = \alpha and right angle at B.
\cos \alpha = \frac{3}{5}=\frac{B}H{}

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=(3)^{2}+(BC)^{2} \\ &25^{2}=9^2+(BC)^2 \\ & (BC)^2 =16 \end{aligned}
BC=\pm 4
BC= 4
\therefore \sin \alpha =\frac{4}{5}
Let, in \Delta PQR, angle RPQ = \beta and right angle at B.
\sin \beta = \frac{5}{13}=\frac{P}H{}

\begin{aligned} &\\ &(PR)^{2}=(PQ)^{2}+(QR)^{2} \\ &(13)^{2}=(PQ)^{2} + 5^2\\ &169=(PQ)^2+25\\ & (PQ)^2 =144 \end{aligned}
PQ=\pm 12 [we will ignore the -ve sign because PQ is a length and it can’t be -ve]
PQ= 12
\therefore \cos \beta =\frac{12}{13}
Now, \begin{aligned} \sin (\alpha+\beta) &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \\ &=\frac{4}{5} \times \frac{12}{13}+\frac{3}{5} \times \frac{5}{13} \\ &=\frac{48}{65}+\frac{15}{65} \\ \sin (\alpha+\beta) &=\frac{63}{65}=\mathrm{RHS} \end{aligned}
Hence proved.

Inverse trigromtery functions exercise 3.8 question 3 (i)

Answer:

x= \pm \sqrt{\frac{35}{36}}
Hint:
We know the value of\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have to solve \cos \left (\sin ^{-1} x \right )= \frac{1}{6} to find the value of x.
Solution:
Here, we have given \sin ^{-1} x in place of \cos ^{-1} x. So, we convert \sin ^{-1} x into \cos ^{-1} x
Let’s suppose that,
\sin ^{-1} x =\alpha
\sin \alpha =x
We use the identity,
\begin{aligned} &\sin ^{2} \alpha+\cos ^{2} \alpha=1 \\ &x^{2}+\cos ^{2} \alpha=1 \\ &\cos ^{2} \alpha=1-x^{2} \end{aligned}
\cos \alpha =\pm \sqrt{1-x^2} [But we can eliminate negative value of \cos \alpha , because in RHS of the question only positive value is given.]
\cos \alpha =\sqrt{1-x^2} …(i)
Now,
\begin{aligned} &\cos \left(\sin ^{-1} x\right)=\frac{1}{6} \\ &\cos \alpha=\frac{1}{6} \end{aligned} .....(ii)

From equation (i) and (ii) we get,
\sqrt{1-x^2}=\frac{1}{6}
Squaring on both sides,
\begin{aligned} &\left(\sqrt{1-x^{2}}\right)^{2}=\left(\frac{1}{6}\right)^{2} \\ &1-x^{2}=\frac{1}{36} \\ &x^{2}=1-\frac{1}{36} \\ &x^{2}=\frac{35}{36} \\ &x=\pm \sqrt{\frac{35}{36}} \end{aligned}
Therefore, the answer is x= \pm \sqrt{\frac{35}{36}} .

Inverse trigromtery functions exercise 3.8 question 3 (ii)

Answer:

x=\pm \frac{1}{\sqrt{2}}
Hint:
As we know that the value of\operatorname{cos}\left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]
Given:
We have to solve \cos \left(2 \sin ^{-1}(-x)\right)=0 to find the value of x.
Solution:
So, we convert 2\sin ^{-1}(x) into \cos ^{-1}(x)
\cos \left(2 \sin ^{-1}(-x)\right)=\cos \left(-2 \sin ^{-1} x\right) [\because \sin ^{-1}(-x)=-\sin x]
=\cos \left(-2 \sin ^{-1} x\right) …(i) [ as, \cos is even function, so \cos(-x)= \cos x]
Let’s suppose,

\sin ^{-1} x =\alpha …(ii)
\sin \alpha =x …(iii)


Given,
\cos \left(2 \sin ^{-1}(-x)\right)=0
\cos \left(2 \sin ^{-1}(-x)\right)=0 … [ from eqn(i) ]
\cos 2 \alpha =0 …[ using (ii) ]
1-2 \sin ^2 \alpha =0 [\cos 2 \alpha =1-2 \sin ^2 \alpha ]
\begin{aligned} &1-2 x^{2}=0 \\ &2 x^{2}=1 \\ &x^{2}=\frac{1}{2} \\ &x=\pm \frac{1}{\sqrt{2}} \end{aligned}

Therefore, the answer is x=\pm \frac{1}{\sqrt{2}}

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Chapter-wise RD Sharma Class 12 Solutions

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