Question

Prove the following trigonometric identities:

$\frac{secA-tanA}{secA+tanA}=\frac{co{s}^{2}A}{(1+sinA{)}^2}$

Answer 1

We know that $se{c}^{2}A-ta{n}^{2}A=1$

$LHS=\frac{secA-tanA}{secA+tanA}$

Multiply both denominator and numerator with $secA+tanA$

$=\frac{(secA-tanA)(secA+tanA)}{(secA+tanA)(secA+tanA)}$

$=\frac{se{c}^{2}A-ta{n}^{2}A}{(secA+tanA{)}^2}$

$=\frac{1}{(se{c}^{2}A+ta{n}^{2}A+2secAtanA)}$

$=\frac{1}{(\frac{1}{co{s}^{2}A}+\frac{si{n}^{2}A}{co{s}^{2}A}+2\times \frac{1}{cosA}\times \frac{sinA}{cosA})}$

$=\frac{1}{(\frac{1}{co{s}^{2}A}+\frac{si{n}^{2}A}{co{s}^{2}A}+\frac{2\times sinA}{co{s}^{2}A})}$

$=\frac{1}{\left(\frac{1+si{n}^{2}A+2sinA}{co{s}^{2}A}\right)}$

$=\frac{co{s}^{2}A}{1+si{n}^{2}A+2sinA}$

$=\frac{co{s}^{2}A}{(1+sinA{)}^2}$

$=RHS$