Question

Prove that $2{\cos }^{2}Q-1=\frac{1-{\tan }^{2}Q}{1+{\tan }^{2}Q}$.

Answer 1

$2{\cos }^{2}Q-1=\frac{1-{\tan }^{2}Q}{1+{\tan }^{2}Q}$

Taking RHS

$=\frac{1-{\tan }^{2}Q}{1+{\tan }^{2}Q}$

$=\frac{1-\frac{{\sin }^{2}Q}{{\cos }^{2}Q}}{1+\frac{{\sin }^{2}Q}{{\cos }^{2}Q}}$$(\tan Q=\frac{\sin Q}{\cos Q})$

$=\frac{\left(\frac{{\cos }^{2}Q-{\sin }^{2}Q}{{\cos }^{2}Q}\right)}{\left(\frac{{\cos }^{2}Q+{\sin }^{2}Q}{{\cos }^{2}Q}\right)}$

$=\frac{{\cos }^{2}Q-{\sin }^{2}Q}{{\cos }^{2}Q}\times \frac{{\cos }^{2}Q}{{\cos }^{2}Q+{\sin }^{2}Q}$

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

$=co{s}^2-(1-co{s}^{2}Q)$$({\sin }^{2}Q=1-co{s}^{2}Q)$

$=2co{s}^{2}Q-1=$ L.H.S

so, $2{\cos }^{2}Q-1=\frac{1-{\tan }^{2}Q}{1+{\tan }^{2}Q}$