Question

Find the value of $\frac{\text{cosec}38}{\sec 52}+\frac{2}{\sqrt{3}}\tan 38.\tan 60.\tan 52-3({\sin }^{2}32+{\sin }^{2}58)$

Answer 1

$\frac{\text{cosec}38}{\sec 52}+\frac{2}{\sqrt{3}}\tan 38\times \tan 60\times \tan 52-3({\sin }^{2}32+{\sin }^{2}58)$
$=\frac{\text{cosec}38}{\text{cosec}(90-52)}+\frac{2}{\sqrt{3}}\tan 38.\sqrt{3}.\cot (90-52)-3({\sin }^{2}32+{\cos }^2(90-58))$
$=\frac{\text{cosec}38}{\text{cosec}38}+2\tan 38.\cot 38-3({\sin }^{2}32+{\cos }^{2}32)$
$=1+2-3\left(1\right)$
$=0$