Question

prove that √1-sinA/1+sinA=secA-tanA

Answer 1

√1-sinA/1+sinA=secA-tanA
L.H.S
√1-sinA/1+sinA
multiple by 1-SinA
√1-sinA/1+sinA×1-SinA/1-SinA
√(1-SinA)^2/√1-(Sin^2A)
1-SinA/√Cos^2
1-SinA/CosA
1/CosA-SinA/CosA
SecA-TanA

Hence Proved

Answer 2

$\sqrt{\dfrac{1-\sin A}{1+\sin A} }=\sec A-\tan A$, proved.

Step-by-step explanation:

We have,

L.H.S. = $\sqrt{\dfrac{1-\sin A}{1+\sin A} }$

Rationalising numerator and denominator by $\sqrt{1-\sin A}$, we get

$= \sqrt{\dfrac{1-\sin A}{1+\sin A}\times \dfrac{1-\sin A}{1-\sin A}}$

$= \sqrt{\dfrac{(1-\sin A)^{2} }{1- \sin^{2} A}$

$= \sqrt{\dfrac{(1-\sin A)^{2} }{\cos^{2} A}$

[ ∵ $1- \sin^{2} A}=\cos^{2} A$ ]

$= \dfrac{(1-\sin A) }{\cos A}$

$= \dfrac{1}{\cos A} -\dfrac{\sin A}{\cos A}$

[ ∵$\dfrac{1}{\cos A} =\sec A$ and $\dfrac{\sin A}{\cos A} =\tan A$ ]

= $\sec A-\tan A$

= R.H.S.

Hence, it is proved.