The value of cot 18° $$\left( {{\text{cot 7}}{{\text{2}}^ \circ }{\text{.co}}{{\text{s}}^2}{{22}^ \circ } + \frac{1}{{{\text{tan 7}}{{\text{2}}^ \circ }.{\text{se}}{{\text{c}}^2}{{68}^ \circ }}}} \right)$$      is?

A. 1

B. $$\sqrt 2 $$

C. 3

D. $$\frac{1}{{\sqrt 3 }}$$

Answer: Option A

Solution(By Examveda Team)

$$\cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \frac{1}{{\tan {{72}^ \circ }.{{\sec }^2}{{68}^ \circ }}}} \right)$$
$$ \Rightarrow \cot {18^ \circ }\left( {\cot {{72}^ \circ }.{{\cos }^2}{{22}^ \circ } + \cot {{72}^ \circ }.{{\cos }^2}{{68}^ \circ }} \right) $$
$$ \Rightarrow \cot {18^ \circ }.\cot {72^ \circ }\left( {{{\cos }^2}{{22}^ \circ } + {{\cos }^2}{{68}^ \circ }} \right) $$
We know that, cotA.cotB = 1 and cos²A + cos²B = 1( when A + B = 90° )
= 1 × 1 = 1