Question

If $-\frac{\pi }{2}<x<-\frac{3\pi }{2}$, then the value of $\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right\}$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B

$-\frac{1}{2}$

Explanation for the correct option

Differentiation of the required trigonometric functions required for the solution are as follows,

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left({\tan }^{-1}\left(x\right)\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}},\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{sin}\left(x\right)\mathbf{=}\mathbf{cos}\left(x\right),\text{and}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{cos}\left(x\right)\mathbf{=}\mathbf{-}\mathbf{sin}\left(x\right)$

Also, the quotient rule of the differentiation is $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(\frac{u}{v}\right)\mathbf{=}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{v}\mathbf{-}\mathbf{u}\mathbf{v}\mathbf{\text{'}}}{{\mathbf{v}}^{\mathbf{2}}}$

Now, the differentiation of the given will be,

$\begin{array}{rcl}\therefore \frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\right\}& =& \frac{1}{1+{\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)}^2}\frac{d}{dx}\left(\frac{\cos \left(x\right)}{1+\sin \left(x\right)}\right)\\ & =& \frac{1}{1+\frac{{\cos }^2\left(x\right)}{{\left(1+\sin \left(x\right)\right)}^2}}\left(\frac{-\sin \left(x\right)\left(1+\sin \left(x\right)\right)-\cos \left(x\right)\left(\cos \left(x\right)\right)}{{\left(1+\sin \left(x\right)\right)}^2}\right)\\ & =& \frac{1}{{\displaystyle \frac{{\left(1+\sin \left(x\right)\right)}^2+{\cos }^2\left(x\right)}{{\left(1+\sin \left(x\right)\right)}^2}}}\left(\frac{-\sin \left(x\right)-{\sin }^2\left(x\right)-{\cos }^2\left(x\right)}{{\left(1+\sin \left(x\right)\right)}^2}\right)\\ & =& \frac{{\left(1+\sin \left(x\right)\right)}^2}{{\left(1+\sin \left(x\right)\right)}^2+{\cos }^2\left(x\right)}\left(\frac{-\sin \left(x\right)-{\sin }^2\left(x\right)-{\cos }^2\left(x\right)}{{\left(1+\sin \left(x\right)\right)}^2}\right)\\ & =& \frac{-\sin \left(x\right)-\left({\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right)}{1+{\sin }^2\left(x\right)+2\sin \left(x\right)+{\cos }^2\left(x\right)}\\ & =& \frac{-1-\sin \left(x\right)}{1+1+2\sin \left(x\right)}\\ & =& \frac{-\left(1+\sin \left(x\right)\right)}{2\left(1+\sin \left(x\right)\right)}\\ & =& -\frac{1}{2}\end{array}$

Hence, the correct option is (B).