Question

If $f\left(x\right)=\frac{1+\sin ({90}^{\circ }+2x)-\cos ({270}^{\circ }+2x)}{\cos ({360}^{\circ }+4x)}$ then value of $\tan {30}^{\circ }f\left({30}^{\circ }\right)+\tan {45}^{\circ }f\left({45}^{\circ }\right)$ is

• A. $0$
• B. $1-\sqrt{3}$
• C. $3\sqrt{3}-1$
• D. $\sqrt{3}-1$

Answer 1

The correct option is B $1-\sqrt{3}$

$f\left(x\right)=\frac{1+\sin ({90}^{\circ }+2x)-\cos ({270}^{\circ }+2x)}{\cos ({360}^{\circ }+4x)}$

$f\left(x\right)=\frac{1+\cos 2x-\sin 2x}{\cos 4x}$

$f\left({30}^{\circ }\right)=\frac{1-\sin (2\times {30}^{\circ })+\cos (2\times {30}^{\circ })}{\cos (4\times {30}^{\circ })}=\sqrt{3}-3$
$f\left({45}^{\circ }\right)=\frac{1-\sin (2\times {45}^{\circ })+\cos (2\times {45}^{\circ })}{\cos (4\times {45}^{\circ })}=0$

Now, $\tan {30}^{\circ }f\left({30}^{\circ }\right)+\tan {45}^{\circ }f\left({45}^{\circ }\right)$
$=\frac{1}{\sqrt{3}}\times (\sqrt{3}-3)+(1\times 0)=1-\sqrt{3}$