Question

Find $\frac{dy}{dx}$, if $y={\tan }^{-1}\left(\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right),\frac{a}{b}\tan x>-1$

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $-2$

$\frac{dy}{dx}$ , if $y=ta{n}^{-1}\left(\frac{acosx-bsinx}{bCosx+asinx}\right)$

$\frac{dy}{dx}=\frac{1}{1+{\left(\frac{acosx-bsinx}{bcosx+asinx}\right)}^2}\frac{(bcosx+asinx)(-asinx-bcosx)-(acosx-bsinx)(-bsinx+acosx)}{{(bcosx+asinx)}^2}$

$=\frac{{(bcosx+asinx)}^2}{{(bcosx+asinx)}^2+{(acosx-bsinx)}^2}$

$=\frac{-absinx.cosx-b^{2}co{s}^{2}x-a^{2}si{n}^{2}x-absinx.cosx-(-absinx.cosx+a^{2}co{s}^{2}x+b^{2}si{n}^{2}x-absinx)bx)}{{(bcosx+ssinx)}^2}$

$\frac{-2absinx.cosx-a^2-b^2+2absinx.cosx-a^2-b^2}{b^{2}cos{x}^2+a^{2}Si{n}^{2}x+2absinx.cosx+a^{2}Co{s}^{2}x+b^{2}si{n}^{2}x-2absinx.cosx}$

$=\frac{-2(a^2+b^2)}{(b^2+a^2)}$

$=-2$