Question

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

• A. tan${46}^{\circ }$
• B. cot${44}^{\circ }$
• C. tan${48}^{\circ }$
• D. cot${48}^{\circ }$

Answer 1

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

$\frac{3+\cot {76}^{\circ }\cot {16}^{\circ }}{\cot {76}^{\circ }+\cot {16}^{\circ }}$

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

$=\frac{2\sin {76}^{\circ }\sin {16}^{\circ }+\sin {76}^{\circ }\sin {16}^{\circ }+\cos {76}^{\circ }\cos {16}^{\circ }}{\sin {16}^{\circ }\cos {76}^{\circ }+\cos {16}^{\circ }\sin {76}^{\circ }}$

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

$=\frac{(\cos {60}^{\circ }-\cos {92}^{\circ })+\cos {60}^{\circ }}{\sin {92}^{\circ }}$

$=\frac{2\cos {60}^{\circ }-\cos {92}^{\circ }}{\sin {92}^{\circ }}$

$=\frac{2\frac{1}{2}-\cos {92}^{\circ }}{\sin {92}^{\circ }}$

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

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

$=\tan {46}^{\circ }$