Question

Find the value of the following expression.
$\frac{\tan {82}^{\circ }}{\cot 8^{\circ }}+\frac{cosec{38}^{\circ }}{\sec {52}^{\circ }}+\frac{2\sin {35}^{\circ }\sec {55}^{\circ }}{{\sin }^2{50}^{\circ }+{\sin }^{2}40}$

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is A

4

Given:
$\frac{\tan {82}^{\circ }}{\cot 8^{\circ }}+\frac{cosec{38}^{\circ }}{\sec {52}^{\circ }}+\frac{2\sin {35}^{\circ }\sec {55}^{\circ }}{{\sin }^2{50}^{\circ }+{\sin }^{2}40}$

$=\frac{\tan ({90}^{\circ }-8^{\circ })}{\cot 8^{\circ }}+\frac{cosec(90-{52}^{\circ })}{\sec {52}^{\circ }}+\frac{2\sin {35}^{\circ }\sec ({90}^{\circ }-{35}^{\circ })}{{\sin }^2{50}^{\circ }+{\sin }^2({90}^{\circ }-{50}^{\circ })}$

We know that,
$tan({90}^o-\theta )=cot\theta ,cosec({90}^o-\theta )=sec\theta sec({90}^o-\theta )=cosec\theta sin({90}^o-\theta )=cos\theta$

On applying these equations we get,
$=\frac{\cot 8^{\circ }}{\cot 8^{\circ }}+\frac{\sec {52}^{\circ }}{\sec {52}^{\circ }}+\frac{2\sin {35}^{\circ }cosec{35}^{\circ }}{{\sin }^2{50}^{\circ }+{\cos }^2{50}^{\circ }}$

$=1+1+\frac{2\left(1\right)}{1}.....(\because sin\theta =\frac{1}{cosec\theta }\text{and}si{n}^2\theta +co{s}^2\theta =1)$

$=1+1+2$

$=4$