#### Provide solution for RD Sharma maths class 12  Chapter Inverse trigromtery functions exercise 3.8 question 3 sub question (i)

$x= \pm \sqrt{\frac{35}{36}}$

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 $x= \pm \sqrt{\frac{35}{36}}$ .