Question

cos A/secA-tanA=1+sinA​

Answer 1

Answer:

hence proved hope it helps you

Answer 2

Answer:

Given:

$\frac{cosA}{secA-tanA}=1+sinA$

We know that cosA=1/SinA, secA=1/cosA, tanA=sinA/cosA. Thus, we can rewrite the equation as:

$\frac{cosA}{ \frac{1}{cosA} -\frac{sinA}{cosA} }=1+sinA$

$\frac{cosA}{\frac{1-sinA}{cosA} }=1+sinA$

Numerator's numerator goes to the denominator... So:

$\frac{cosA*cosA}{1-sinA}=1+sinA$

$\frac{cos^{2}A }{1-sinA}=1+sinA$

Here, cos²A=1-sin²A. Thus, we can rewrite the equation as:

$\frac{1-sin^{2}A }{1-sinA}=1+sinA$

Here, 1-sin²A is in the form a²-b²=(a+b)(a-b) Thus, 1-sin²A=(1+sinA)(1-sinA). Thus, we can rewrite the equation as:

$\frac{(1-sinA)(1+sinA) }{1-sinA}=1+sinA$

$1+sinA=1+sinA$

Hence, RHS=LHS, thus proved.

HOPE THIS HELPS :D.