Question

Prove that $(1+\sec {20}^o+\cot {70}^o)(1-cosec{20}^o+\tan {70}^o)=2$.

Answer 1

Now,

$(1+\sec {20}^o+\cot {70}^o)(1-cosec{20}^o+\tan {70}^o)$

$=(1+\sec {20}^o+\tan {20}^o)(1-cosec{20}^o+\cot {20}^o)$ [ $\tan {70}^o=\cot {20}^o$ and $\cot {70}^o=\tan {20}^o$]

$=(1+\frac{1}{\cos {20}^o}+\frac{\sin {20}^o}{\cos {20}^o})(1-\frac{1}{\sin {20}^o}+\frac{\cos {20}^o}{\sin {20}^o})$

$=\left(\frac{\cos {20}^o+\sin {20}^o+1}{\cos {20}^o}\right)\left(\frac{\sin {20}^o+\cos {20}^o-1}{\sin {20}^o}\right)$

$=\left(\frac{(\cos {20}^o+\sin {20}^o{)}^2+1}{\sin {20}^o.\cos {20}^o}\right)$

$=\left(\frac{2\sin {20}^o.\cos {20}^o}{\sin {20}^o.\cos {20}^o}\right)$ [Since ${\sin }^2{20}^o+{\cos }^2{20}^o=1$]

$=2$.