Question

Without using trigonometric tables, evaluate
$\frac{\sec ({90}^o-\theta )}{\csc \theta }+\sqrt{3}[\tan {10}^o.\tan {40}^o.\tan {50}^o.\tan {80}^o]\times \frac{2{\sin }^2{45}^o}{{\sec }^2{22}^o-{\cot }^2{68}^o}$

Answer 1

We have,

$\frac{\sec ({90}^o-\theta )}{\csc \theta }+\sqrt{3}[\tan {10}^o.\tan {40}^o.\tan {50}^o.\tan {80}^o]\times \frac{2{\sin }^2{45}^o}{{\sec }^2{22}^o-{\cot }^2{68}^o}$

$\Rightarrow \frac{\csc \theta }{\csc \theta }+\sqrt{3}\left[\tan {10}^o\tan {40}^o\cot {40}^o\cot {10}^o\right]\times \frac{2{\sin }^2{45}^o}{{\sec }^2{22}^o-{\cot }^2({90}^o-22)}$

$\Rightarrow 1+\sqrt{3}\left[\tan {10}^o\tan {40}^o\frac{1}{\tan {40}^o}\frac{1}{\tan {10}^o}\right]\times \frac{2{\sin }^2{45}^o}{{\sec }^2{22}^o-{\tan }^2{22}^o}\therefore {\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow 1+\sqrt{3}\times 1\times \frac{2{\sin }^2{45}^o}{1}$

$\Rightarrow 1+\sqrt{3}\times 2{\sin }^2{45}^o$

$\Rightarrow 1+2\sqrt{3}\times \frac{1}{2}$

$\Rightarrow 1+\sqrt{3}$

Hence, this is the answer.