Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

${\mathrm{Sin}}^{-1}(x)+{\mathrm{Cos}}^{-1}(x)=\frac{\pi }{2}$, we have

$$x-\frac{x^2}{2}+\frac{x^3}{4}+\cdots =x^2-\frac{x^4}{2}+\frac{x^6}{4}+\dots$$

Both sides are infinite geometric progressions. Thus,

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

$$\frac{2x}{2+x}=\frac{2x^2}{2+x^2}$$

If $x\ne 0$, then $\frac{1}{2+x}=\frac{x}{2+x^2}$, so $2+x^2=2x+x^2$, which gives $2x=2$, so $x=1$.

Final Answer: 1