RD Sharma Class 12 Exercise 3.8 Inverse Trigonometry functions Solutions Maths

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.

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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 $\sin ^{-1} x$ we 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 $\sin ^{-1} \frac{3}{5}= \alpha$ and $\cot ^{-1} \frac{3}{2}=\beta$ .
$\sin \alpha = \frac{3}{5} \; and\; \cot \beta =\frac{3}{2}$
$\cos\left ( \alpha +\beta \right )= \cos \alpha \cos \beta -\sin \alpha \sin \beta$
Here, we only have $\sin \alpha$ and $\cos \beta$ .So we have to find $\cos \alpha$ and $\sin \beta$ .
$\sin \alpha =\frac{3}{5}=\frac{B}{H}$
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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4. Is it enough if I work out the sums for chapter 3 given in the RD Sharma book?

It is more than enough to work out the sums in the RD Sharma Class 12 Chapter 3 Exercise 3.8 solutions for a deeper understanding of its mathematical concepts.

5. Are the solutions provided in RD Sharma Class 12th exercise 3.8 easy to understand?

With the solutions of educational experts, RD Sharma Class 12th exercise 3.8 solution consists of many methods of solving a mathematical function. Hence, both the toppers and the average students can find it easier to choose the way they understand and go ahead with it.