Question

whole root 1- sinA/1+sinA=secA-tanA​

Answer 1

QUESTION:

Prove that

$\sqrt {\dfrac{1- sinA}{1+sinA}} = \sec A- \tan A$

GIVEN:

$\sqrt {\dfrac{1- sinA}{1+sinA}} = \sec A- \tan A$

TO PROVE:

$\sqrt {\dfrac{1- sinA}{1+sinA}} = \sec A- \tan A$

EXPLANATION:

$\sqrt {\dfrac{1- sinA}{1+sinA}} = \sec A- \tan A$

$Take \ L.H.S \ as \ \sqrt {\dfrac{1- sinA}{1+sinA}}$

$Multiply \ by \ \sqrt {\dfrac{1- sinA}{1 - sinA}} \ on \ both \ numerator$$and \: denominator$

$\boxed{\large{\bold{(A + B)(A - B) = A^2 - B^2}}}$

We will get

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

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

$\sf\boxed{\huge{\bold{ \sqrt{ {x}^{2} } = x}}}$

$\sf\boxed{\huge{\bold{1 - sin^{2} A = {cos}^{2} A}}}$

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

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

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

$\sec A - \tan A$

HENCE PROVED.