Question

Solve it plz...
$\frac{\sec(a) - 1}{\sec(a) + 1} =( \frac{ { \sin(a) } }{1 + \cos(a) } ) ^{2}$
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Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{secA-1}{secA+1}=\left(\dfrac{sinA}{1+cosA}\right)^2$

How to Do :-

First of all we need to covert 'secA' in terms of 'cosA' and we need to simplify it. Next by applying a trigonometric identity we can prove that L.H.S = R.H.S.

Formula Required :-

secA = 1/cosA

sin²A + cos²A = 1

Solution :-

Taking L.H.S :-

= $\dfrac{secA-1}{secA+1}$

= $\dfrac{\dfrac{1}{cosA}-1}}{\dfrac{1}{cosA}+1}$

[∴ secA = 1/cosA]

$=\dfrac{\dfrac{1-cosA}{cosA}}{\dfrac{1+cosA}{cosA}}$

$=\dfrac{1-cosA}{cosA}\times \dfrac{cosA}{1+cosA}$

$=\dfrac{1-cosA}{1+cosA}$

Multiplying and dividing with '1 + cosA' :-

$=\dfrac{1-cosA}{1+cosA}\times \dfrac{1+cosA}{1+cosA}$

$=\dfrac{1^2-cos^2A}{(1+cosA)^2}$

$=\dfrac{1-cos^2A}{(1+cosA)^2}$

$=\dfrac{sin^2A}{(1+cosA)^2}$

[∴ 1 - cos²A = 1]

$=\left(\dfrac{sinA}{1+cosA}\right)^2$

= R.H.S

Hence Proved.