Question

Prove that:

$\frac{\sec A-\tan A}{\sec A+\tan A}={\left(\frac{\cos A}{1+\sin A}\right)}^2$.

Answer 1

$L.H.S=\frac{secA-tanA}{secA+tanA}$

Rationalizing the denominator by multiply and dividing with sec A+ tan A.

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

$=\frac{1}{se{c}^{2}A+ta{n}^{2}A+2.secA.tanA}$ $[\because se{c}^{2}A-ta{n}^{2}A=1]$

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

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

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

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

$=R.H.S$