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?
A. 2
B. 1
C. 0
D. -1
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \frac{{2{{\tan }^2}{{30}^ \circ }}}{{1 - {{\tan }^2}{{30}^ \circ }}} + {\sec ^2}{45^ \circ } - {\sec ^2}{0^ \circ } = x\sec {60^ \circ } \cr & \Rightarrow \frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}} + {\left( {\sqrt 2 } \right)^2} - 1 = x \times 2 \cr & \Rightarrow \frac{{2 \times \frac{1}{3}}}{{1 - \frac{1}{3}}} + 2 - 1 = 2x \cr & \Rightarrow \left( {\frac{2}{3} \times \frac{3}{2}} \right) + 2 - 1 = 2x \cr & \Rightarrow 2 = x \times 2 \cr & \Rightarrow x = 1 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
Join The Discussion