# If, then find the value of x.

Question:

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then find the value of $x$.

Solution:

$\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$

$\Rightarrow \sin \left(\sin ^{-1} \frac{1}{5}\right) \cos \left(\cos ^{-1} x\right)+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1$

$[\sin (A+B)=\sin A \cos B+\cos A \sin B]$

$\Rightarrow \frac{1}{5} \times x+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1$

$\Rightarrow \frac{x}{5}+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1$....(1)

Now, let $\sin ^{-1} \frac{1}{5}=y$.

Then, $\sin y=\frac{1}{5} \Rightarrow \cos y=\sqrt{1-\left(\frac{1}{5}\right)^{2}}=\frac{2 \sqrt{6}}{5} \Rightarrow y=\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$.

$\therefore \sin ^{-1} \frac{1}{5}=\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$....(2)

Let $\cos ^{-1} x=z$.

Then, $\cos z=x \Rightarrow \sin z=\sqrt{1-x^{2}} \Rightarrow z=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)$

$\therefore \cos ^{-1} x=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)$....(3)

From $(1),(2)$, and $(3)$ we have:

$\frac{x}{5}+\cos \left(\cos ^{-1} \frac{2 \sqrt{6}}{5}\right) \cdot \sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)=1$

$\Rightarrow \frac{x}{5}+\frac{2 \sqrt{6}}{5} \cdot \sqrt{1-x^{2}}=1$

$\Rightarrow x+2 \sqrt{6} \sqrt{1-x^{2}}=5$

$\Rightarrow 2 \sqrt{6} \sqrt{1-x^{2}}=5-x$

On squaring both sides, we get:

$(4)(6)\left(1-x^{2}\right)=25+x^{2}-10 x$

$\Rightarrow 24-24 x^{2}=25+x^{2}-10 x$

$\Rightarrow 25 x^{2}-10 x+1=0$

$\Rightarrow(5 x-1)^{2}=0$

$\Rightarrow(5 x-1)=0$

$\Rightarrow x=\frac{1}{5}$

Hence, the value of $x$ is $\frac{1}{5}$.