Question

Prove that : root (1 + sin theta / 1 - sin theta) + (1-sin theta / 1 + sin theta) = 2 sec theta

Answer 1

Proved the above question .

Answer 2

To prove: $\sqrt{\dfrac{1+\sin \theta}{1-\sin \theta} } +\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } =2 \sec \theta$

Step-by-step explanation:

Consider : L.H.S. we have

$\sqrt{\dfrac{1+\sin \theta}{1-\sin \theta} } +\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } \\\\=\sqrt{\dfrac{1+\sin \theta}{1-\sin \theta} \times \dfrac{1+\sin \theta}{1+\sin \theta} } +\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} \times \dfrac{1-\sin \theta}{1-\sin \theta}} \\\\=\sqrt{\dfrac{(1+ \sin \theta)^2}{1^2-\sin^2 \theta} } +\sqrt{\dfrac{(1- \sin \theta)^2}{1^2-\sin^2 \theta} } \\\\\text {As we know }\ sin^2A+\cos^2A=1$

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

Which is equal to R.H.S.

Hence proved the required result