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