Question

$\frac{1}{2}\sin \left(2x\right)(1+{\cot }^2\left(x\right))$ is equal to

• A. $\tan \left(x\right)$
• B. $\sin \left(x\right)$
• C. $\cos \left(x\right)$
• D. $\cot \left(x\right)$
• E. $\sec \left(x\right)$

Answer 1

The correct option is D $\cot \left(x\right)$

The given identity can be solved as:

$\left(\frac{1}{2}\right)\left(\sin 2x)\right(1+{\cot }^{2}x)$

$=\left(\frac{1}{2}\right)\left(2\sin x\cos x\right)\left(cose{c}^{2}x\right)$ .... (because $\sin 2x=2\sin x\cos x$ and $1+{\cot }^{2}x=cose{c}^{2}x$)

$=\left(\sin x\cos x\right)\left(\frac{1}{{\sin }^{2}x}\right)$ ...... (because $cose{c}^{2}x=\frac{1}{{\sin }^{2}x}$)

$=\frac{\cos x}{\sin x}$

$=\cot x$

Hence, $\left(\frac{1}{2}\right)\left(\sin 2x)\right(1+{\cot }^{2}x)=\cot x$.