Prove the identity: $\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=2sec\theta$.

Given: $\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=2sec\theta$.

To do: To prove that $L.H.S.=R.H.S.$

Solution: 

$L.H.S.=\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}$

$=\sqrt{\frac{( 1+sin\theta)( 1+sin\theta)}{( 1-sin\theta)( 1+sin\theta)}}+\sqrt{\frac{( 1-sin\theta)( 1-sin\theta)}{( 1+sin\theta)( 1-sin\theta)}}$

$=\sqrt{\frac{( 1+sin\theta)^{2}}{1-sin^{2}\theta}}+\sqrt{\frac{( 1-sin\theta)^{2}}{1-sin^{2}\theta}}$

$=\sqrt{\frac{( 1+sin\theta)^{2}}{cos^{2}\theta}}+\sqrt{\frac{( 1-sin\theta)^{2}}{cos^{2}\theta}}$

$=\sqrt{( \frac{1+sin\theta}{cos\theta})^{2}}+\sqrt{( \frac{1-sin\theta}{cos\theta})^{2}}$

$=( \frac{1+sin\theta}{cos\theta})+( \frac{1-sin\theta}{cos\theta})$

$=\frac{1+sin\theta+1-sin\theta}{cos\theta}$

$=\frac{2}{cos\theta}$

$=2sec\theta$

$=R.H.S.$

Hence, the identity has been proved that $\sqrt{\frac{1+sin\theta}{1-sin\theta}}+\sqrt{\frac{1-sin\theta}{1+sin\theta}}=2sec\theta$.

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Updated on: 10-Oct-2022

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