Question

If $x^2={\sin }^2{30}^o+4{\cot }^2{45}^o-{\sec }^2{60}^o$, then find the value of $x$.

Answer 1

Given,

$x^2={\sin }^2{30}^o+4{\cot }^2{45}^o-{\sec }^2{60}^o$

or, $x^2=\frac{1}{4}+4.1-4$ [ Using the values of $\sin {30}^o,\cot {45}^o,\sec {60}^o$]

or, $x^2=\frac{1}{4}$

or, $x=\pm \frac{1}{2}$