Question

Prove the following trigonometric identities:

$\frac{ta{n}^{2}A}{1+ta{n}^{2}A}+\frac{co{t}^{2}A}{1+co{t}^{2}A}=1$

Answer 1

$LHS=\frac{ta{n}^{2}A}{1+ta{n}^{2}A}+\frac{co{t}^{2}A}{1+co{t}^{2}A}$

$=\frac{ta{n}^{2}A}{se{c}^{2}A}+\frac{co{t}^{2}A}{cose{c}^{2}A}$

$=\frac{si{n}^{2}A}{co{s}^{2}A}\times co{s}^{2}A+\frac{co{s}^{2}A}{si{n}^{2}A}\times si{n}^{2}A$

$=si{n}^{2}A+co{s}^{2}A$

$=1=RHS$