Question

Solve $\sec {50}^o\sec {40}^o+\text{cosec}{40}^o\text{cosec}{50}^o$

Answer 1

$\sec {50}^{\circ }.\sec {40}^{\circ }+cosec{40}^{\circ }.cosec{50}^{\circ }$

$=\frac{1}{\cos 50.\cos 40}+\frac{1}{\sin 50.\sin 40}$

$=\frac{\sin 50.\sin 40+\cos 50.\cos 40}{\sin 50.\cos 50.\sin 40.\cos 40}$

multiply and divide with

$=\frac{2\times 2(\sin 50.\sin 40+\cos 50.\cos 40)}{4\sin 50.\cos 50.\sin 40.\cos 40}$

$=\frac{2(2\sin 50.\sin 40+2\cos 50.\cos 40)}{(2\sin 50.\cos 50)(2\sin 40.\cos 40)}$

$\left[\begin{array}{c}\because 2\cos A.\cos B=\cos (A+B)+\cos (A-B)\\ 2\sin A.\sin B=\cos (A-B)-\cos (A+B)\\ 2\sin A.\cos A=\sin 2A\end{array}\right]$

$=\frac{2\left[\right(\cos (50-40)-\cos (50+40))+(\cos (50+40)+\cos (50-40)\left)\right]}{\sin 100.\sin 80}$

$=\frac{2(\cos 10-\cos 90+\cos 90+\cos 10)}{\sin (90+10).\sin (90-10)}$

$\therefore \sin (90\pm \theta )=\cos \theta$

$=\frac{2\times 2\cos 10}{\cos 10.\cos 10}=\frac{4}{\cos 10}=4\sec {10}^{\circ }$