Question

The value of $\frac{3+\cot {76}^{\circ }\cot {16}^{\circ }}{\cot {76}^{\circ }+\cot {16}^{\circ }}$ is

• A. $\tan {44}^{\circ }$
• B. $\cot {46}^{\circ }$
• C. $\tan 2^{\circ }$
• D. $\tan {46}^{\circ }$

Answer 1

The correct option is D $\tan {46}^{\circ }$

$=\frac{3\tan 76\tan 16+1}{\tan 76+\tan 16}$

$=\frac{1-\tan 76\tan 16}{\tan 76+\tan 16}+2\frac{\left(2\tan 76\tan 16\right)}{\tan 76+\tan 16}$

$=\frac{3\sin 76\sin 16+\cos 76\cos 16}{\sin 76\cos 16+\cos 76\sin 16}$

$=\frac{2\sin 76\sin 16+\sin 76\sin 16+\cos 76\cos 16}{\sin (76+16)}$

$=\frac{\cos (76-16)-\cos (76+16)}{\sin (76+16)}$

$=\frac{2\cos \left({\widetilde{76-16}}^{60}\right)-cos\left({\widetilde{76+16}}^{92}\right)}{\sin \left(\underset{92}{\underset{}{76+16}}\right)}$

$=\frac{1-\cos 92}{\sin 92}=\tan {46}^{\circ }$