If ( ^-1(x√2))= ( ^-1√1-x^2),x (0,1), then the value of x is:

Question

If $\sin ({	an }^{-1}(x\sqrt{2}))=\cot ({\sin }^{-1}\sqrt{1-{x}^{2}}),x\in (0,1)$, then the value of $x$ is:

Accepted Answer

Let us solve

$\sin ({	an }^{-1}(x\sqrt{2}))=\cot ({\sin }^{-1}\sqrt{1-{x}^{2}}),\, x\in (0,1)$

step by step.

Simplify the left-hand side: Let

$	heta ={	an }^{-1}(x\sqrt{2})$

Then

$	an 	heta =x\sqrt{2}$

We know that if $	an 	heta =\frac{}{}$, then we can take

opposite $=x\sqrt{2}$

adjacent $=1$

So hypotenuse is

$\sqrt{1+(x\sqrt{2}{)}^{2}}=\sqrt{1+2{x}^{2}}$

Hence,

$\sin 	heta =\frac{x\sqrt{2}}{\sqrt{1+2{x}^{2}}}$

Therefore,

$\sin ({	an }^{-1}(x\sqrt{2}))=\frac{x\sqrt{2}}{\sqrt{1+2{x}^{2}}}$

Simplify the right-hand side: Let

$\phi ={\sin }^{-1}\sqrt{1-{x}^{2}}$

Then

$\sin \phi =\sqrt{1-{x}^{2}}$

Since $x\in (0,1)$, we have $\sqrt{1-{x}^{2}}\in (0,1)$, so $\phi $ is acute.

Now,

$\cos \phi =\sqrt{1-{\sin }^{2}\phi }=\sqrt{1-(1-{x}^{2})}=\sqrt{{x}^{2}}=x$

because $x>0$.

Therefore,

$\cot \phi =\frac{\cos \phi }{\sin \phi }=\frac{x}{\sqrt{1-{x}^{2}}}$

So,

$\cot ({\sin }^{-1}\sqrt{1-{x}^{2}})=\frac{x}{\sqrt{1-{x}^{2}}}$

Equate both sides: Now the given equation becomes

$\frac{x\sqrt{2}}{\sqrt{1+2{x}^{2}}}=\frac{x}{\sqrt{1-{x}^{2}}}$

Since $x\in (0,1)$, we have $x
e 0$, so divide both sides by $x$:

$\frac{\sqrt{2}}{\sqrt{1+2{x}^{2}}}=\frac{1}{\sqrt{1-{x}^{2}}}$

Cross multiply: $\sqrt{2}\, \sqrt{1-{x}^{2}}=\sqrt{1+2{x}^{2}}$

Now square both sides:

$2(1-{x}^{2})=1+2{x}^{2}$

$2-2{x}^{2}=1+2{x}^{2}$

$1=4{x}^{2}$

${x}^{2}=\frac{1}{4}$

Since $x\in (0,1)$,

$x=\frac{1}{2}$

Final Answer
$\frac{1}{2}$

So, the correct option is Option A.

Solutions · 1