Question

The value of $\frac{\cos {62}^{\circ }}{\sin {52}^{\circ }\sin {66}^{\circ }}+\frac{\cos {52}^{\circ }}{\sin {66}^{\circ }\sin {62}^{\circ }}+\frac{\cos {66}^{\circ }}{\sin {62}^{\circ }\sin {52}^{\circ }}$ is equal to :

Answer 1

$\cos {62}^0=\cos ({180}^0-{118}^0);\cos {52}^0=\cos ({180}^0-{128}^0);\cos {66}^0=\cos ({180}^0-{114}^0)$

$=-\cos \left({118}^0\right)$ $;$ $=-\cos \left({128}^0\right)$ $;$ $=-\cos \left({114}^0\right)$

$=-\cos ({52}^0+{66}^0)$ $;$ $=-\cos ({66}^0+{62}^0)$ $;$ $=-\cos ({62}^0+{52}^0)$

$\therefore$ $\frac{\cos {62}^0}{\sin {52}^0\sin {66}^0}+\frac{\cos {52}^0}{\sin {66}^0\sin {62}^0}+\frac{\cos {66}^0}{\sin {62}^0\sin {52}^0}$

$=-\frac{[\cos {52}^0\cos {66}^0-\sin {52}^0\sin {66}^0]}{\sin {52}^0\sin {66}^0}+\frac{-[\cos {66}^0\cos {62}^0-\sin {66}^0\sin {62}^0]}{\sin {66}^0\sin {62}^0}+\frac{-[\cos {62}^0\cos {52}^0-\sin {62}^0\sin {52}^0]}{\sin {62}^0\sin {52}^0}$

$=[1-\cot {52}^0\cot {66}^0]+[1-\cot {66}^0\cot {62}^0]+[1-\cot {62}^0\cot {52}^0]$

$=3-\cot {52}^0\cot {66}^0-\cot {66}^0\cot {62}^0-\cot {62}^0\cot {52}^0$

Now, ${52}^0+{66}^0+{62}^0={180}^0$

We know if $A+B+C={180}^0$

$cotAcotB+cotBcotC+cotCcotA=1$

$\therefore$ $\frac{\cos {62}^0}{\sin {52}^0\sin {66}^0}+\frac{\cos {52}^0}{\sin {66}^0\sin {62}^0}+\frac{\cos {66}^0}{\sin {52}^0\sin {62}^0}=3-1=2$