Question

$y={\sin }^{-1}\left\{\frac{a\sin x+b\cos x}{\sqrt{a^2+b^2}}\right\}$, then $\frac{dy}{dx}=$

• A. $1$
• B. $-1$
• C. $0$
• D. $1/2$

Answer 1

The correct option is A $1$

Consider the following question.

$y={\sin }^{-1}\left(\frac{a\sin x+b\cos x}{\sqrt{a^2+b^2}}\right)$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(\frac{a\sin x+b\cos x}{\sqrt{a^2+b^2}}\right)}^2}}(a\cos x-b\sin x)$

$=\frac{\sqrt{(a^2+b^2)}}{\sqrt{(a^2+b^2)-{(a\sin x+b\cos x)}^2}}\frac{(a\cos x-b\sin x)}{\sqrt{a^2+b^2}}$

$=\frac{(a\cos x-b\sin x)}{\sqrt{a^2+b^2-a^2{\sin }^{2}x-b^2{\cos }^{2}x-2b^2{a}^2\sin x\cos x}}$

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

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

$=\frac{(a\cos x-b\sin x)}{\sqrt{a^2{\cos }^{2}x+b^2{\sin }^{2}x-2b^2{a}^2\sin x\cos x}}$

$=\frac{(a\cos x-b\sin x)}{(a\cos x-b\sin x)}$

$=1$

Hence, this is the required answer.