Question

Prove question number 5​

Answer 1

$\tt \frac{1-sin\theta}{1+sin\theta} = {(sec\theta + tan\theta)}^{2} \\ \\ \tt \longrightarrow L.H.S. = \frac{1-sin\theta}{1+sin\theta} \\ \\ \tt \longrightarrow \frac{(1-sin\theta)(1-sin\theta)}{(1+sin\theta)(1-sin\theta)} \\ \\ \tt \longrightarrow \frac{{(1-sin\theta)}^{2}}{1 - {sin}^{2}\theta} \\ \\ \tt \longrightarrow \frac{{(1-sin\theta}^{2}}{{cos}^{2}\theta} \\ \\ \tt \longrightarrow {(\frac{1-sin\theta}{cos\theta})}^{2} \\ \\ \tt \longrightarrow {(\frac{1}{cos\theta} + \frac{sin\theta}{cos\theta})}^{2} \\ \\ \tt \longrightarrow {(sec\theta + tan\theta)}^{2} \\ \\ \tt \longrightarrow R.H.S. \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\frac{1-sin\theta}{1+sin\theta} = {(sec\theta + tan\theta)}^{2} }}}}$

Answer 2

Answer:

$\tt \frac{1-sin\theta}{1+sin\theta} = {(sec\theta + tan\theta)}^{2} \\ \\ \tt \longrightarrow L.H.S. = \frac{1-sin\theta}{1+sin\theta} \\ \\ \tt \longrightarrow \frac{(1-sin\theta)(1-sin\theta)}{(1+sin\theta)(1-sin\theta)} \\ \\ \tt \longrightarrow \frac{{(1-sin\theta)}^{2}}{1 - {sin}^{2}\theta} \\ \\ \tt \longrightarrow \frac{{(1-sin\theta}^{2}}{{cos}^{2}\theta} \\ \\ \tt \longrightarrow {(\frac{1-sin\theta}{cos\theta})}^{2} \\ \\ \tt \longrightarrow {(\frac{1}{cos\theta} + \frac{sin\theta}{cos\theta})}^{2} \\ \\ \tt \longrightarrow {(sec\theta + tan\theta)}^{2} \\ \\ \tt \longrightarrow R.H.S. \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\frac{1-sin\theta}{1+sin\theta} = {(sec\theta + tan\theta)}^{2} }}}}$