Question

$\frac{3+\cot {76}^0\cot {16}^0}{\cot {76}^0+\cot {16}^0}$ is equal to

• A. $\tan {16}^0$
• B. $\cot {76}^0$
• C. $\tan {46}^0$
• D. $\cot {44}^0$

Answer 1

Answer: (C), (D)

$\tan {46}^0$
$\cot {44}^0$

Since, $\frac{3+\cot {76}^0\cot {16}^0}{\cot {76}^0+\cot {16}^0}$

$=\frac{3\sin {76}^0\sin {16}^0+\cos {76}^0\cos {16}^0}{\sin ({76}^0+{16}^0)}$

$=\frac{2\sin {76}^0\sin {16}^0+\cos ({76}^0-{16}^0)}{\sin \left({92}^0\right)}$

$=\frac{\cos {60}^0-\cos {92}^0+\cos {60}^0}{\sin {92}^0}$

$=\frac{1-\cos {92}^0}{\sin {92}^0}$

$=\tan {46}^0$ [option (C)]

$=\tan ({90}^0-{44}^0)=\cot {44}^0$ [option (D)]