Question

Evaluate the following:
$\sec {50}^0\sin {40}^0+\cos {40}^0\text{cosec}{50}^0$

Answer 1

To evaluate,

$\sec {50}^0\sin {40}^0+\cos {40}^0\csc {50}^0$

We know that,

$\sec \theta =\frac{1}{\cos \theta }$ , $\csc \theta =\frac{1}{\sin \theta }$

$\therefore \sec {50}^0\sin {40}^0+\cos {40}^0\csc {50}^0$

$=\frac{\sin {40}^0}{\cos {50}^0}+\frac{\cos {40}^0}{\sin {50}^0}$

We have $\sin ({90}^0-\theta )=\cos \theta$

$\sin \left({40}^0\right)=\sin ({90}^0-{50}^0)=\cos \left({50}^0\right)$

$\sin \left({50}^0\right)=\sin ({90}^0-{40}^0)=\cos \left({40}^0\right)$

Therefore,

$=\frac{\sin {40}^0}{\cos {50}^0}+\frac{\cos {40}^0}{\sin {50}^0}$

$=\frac{\cos {50}^0}{\cos {50}^0}+\frac{\cos {40}^0}{\cos {40}^0}$

$=1+1$

$=2$

Hence, $\sec {50}^0\sin {40}^0+\cos {40}^0\csc {50}^0=2$.