Question

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

• A. $\cot {60}^o$
• B. $\tan 2^o$
• C. $\tan {46}^o$
• D. $\cot {45}^o$

Answer 1

Answer: (C)

$\tan {46}^o$

$\frac{3+\cot 76°\cot 16°}{\cot 76°\cot 16°}$

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

$=\frac{2\sin 76°\sin 16°+\sin 76°\sin 16°+\cos 76°\cos 16°}{\cos 76°\sin 16°+cos16°sin76°}$

$=\frac{2\sin 76°\sin 16°+\cos 60°}{\sin (76°+16°)}$

$=\frac{\cos 60°-\cos 92°+\cos 60°}{\sin 92°}$

$=\frac{1-\cos 92°}{\sin 92°}$

$=\frac{2{\sin }^{2}46°}{2\sin 46°\cos 46°}$

$=\tan 46°$