If $$\frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}}$$   + $${\sec ^2}{45^ \circ }$$  - $${\sec ^2}{0^ \circ }$$  = $$x\sec {60^ \circ }{\text{,}}$$   then the value of x is?

Solution (By Examveda Team)

$$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$