Question

Question 5 (vi)
Prove the following identities, where the angles involved are acute angles for which the expressions is defined.
sec A + tan A = $\sqrt{\left(\frac{1+sinA}{1-sinA}\right)}$

Answer 1

$\sqrt{\left(\frac{1+sinA}{1-sinA}\right)}=R.H.S\text{Dividing R.H.S by cos A}:=\sqrt{\left(\frac{\frac{1}{cosA}+\frac{sinA}{cosA}}{\frac{1}{cosA}-\frac{sinA}{cosA}}\right)}=\sqrt{\left(\frac{secA+tanA}{secA-tanA}\right)}=\sqrt{\left(\frac{secA+tanA}{secA-tanA}\right)\times \left(\frac{secA+tanA}{secA+tanA}\right)}=\frac{secA+tanA}{\sqrt{se{c}^{2}A-ta{n}^{2}A}}=\frac{secA+tanA}{1}(\because se{c}^{2}A-ta{n}^{2}A=1)=secA+tanA=L.H.S\Rightarrow L.H.S=R.H.SProved.$