Question

Prove that ${(\tan \theta +\frac{1}{\cos \theta })}^2+{(\tan \theta -\frac{1}{\cos \theta })}^2=2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)$

Answer 1

Formula : $(a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$

LHS is in the form of the above formula

$LHS={(\tan \theta +\frac{1}{\cos \theta })}^2+{(\tan \theta -\frac{1}{\cos \theta })}^2$

$=2({\tan }^2\theta +\frac{1}{{\cos }^2\theta })$

$=2(\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{1}{{\cos }^2\theta })$

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

$=2\left(\frac{1+{\sin }^2\theta }{1-si{n}^2\theta }\right)$

$=RHS$

Hence proved