Question

cos a / 1+ sin a = sec a - tan a​

Answer 1

LHS cosa / 1+sina

cosa/1+sina×1-sina/1-sina

cosa×1-sina /1-sin^2a

cosa ×1-sina /cos^2a

1-sina/cosa

1/cosa -sina/cosa

seca -tana = RHS

Answer 2

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

Step-by-step explanation:

To prove, $\dfrac{\cos A}{1+\sin A}=\sec A-\tan A$

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

Rationalising numerator and denominator, we get

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

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

Using the formula,

$a^{2}-b^{2}=(a+b)(a-b)$

$[tex]=\dfrac{\cos A(1-\sin A)}{\cos^2 A}$

Using the trigonometric formula,

$\cos^2 A=1-\sin^2 A$

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

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

$=\sec A-\tan A$

=R.H.S. proved.