Question

Prove the following identity ​

Answer 1

QUESTION:

Prove the following identity:

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 -sin \ \theta)} = 1 + \dfrac{1}{ sin \ \theta}$

GIVEN:

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 - cos \ \theta)} = 1 + \dfrac{1}{ sin \ \theta}$

TO PROVE:

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 - sin\ \theta)} = 1 + \dfrac{1}{ sin \ \theta}$

EXPLANATION:

$\sf Take\ L.H.S\ as \ \dfrac{cos \ \theta }{ tan \ \theta(1 - sin\ \theta)}$

$\sf Take\ R.H.S \ as\ 1 + \dfrac{1}{ sin \ \theta}$

Take L.H.S:

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 - sin\ \theta)}$

Multiply by 1 + sin θ on both numerator and denominator.

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 - sin\ \theta)} \times \dfrac{(1 + sin\ \theta)}{(1 + sin\ \theta)}$

$\sf \dfrac{cos \ \theta(1 + sin\ \theta) }{ tan \ \theta(1 - sin^{2} \ \theta)}$

$\boxed{ \bold{ \large{ \gray{ 1 - sin^2\ \theta = cos^2\ \theta}}}}$

$\sf \dfrac{cos \ \theta(1 + sin\ \theta) }{ tan \ \theta(cos^{2} \ \theta)}$

$\sf \dfrac{1 + sin\ \theta}{ tan \ \theta(cos \ \theta)}$

$\boxed{ \bold{ \large{ \gray{tan\ \theta = \frac{sin \ \theta}{ cos\ \theta}}}}}$

$\sf \dfrac{1 + sin\ \theta}{ \dfrac {sin\ \theta}{ cos \ \theta}(cos \ \theta)}$

$\sf \dfrac{1 + sin\ \theta}{sin\ \theta}$

$\sf \dfrac{ sin\ \theta}{sin\ \theta} + \dfrac{1}{sin\ \theta}$

$\sf 1+ \dfrac{1}{sin\ \theta}$

Take R.H.S:

$\sf 1 + \dfrac{1}{ sin \ \theta}$

L.H.S = R.H.S

$\sf \dfrac{cos \ \theta }{ tan \ \theta(1 - cos \ \theta)} = 1 + \dfrac{1}{ sin \ \theta}$

HENCE PROVED.

Note : Refer attachment.

Answer 2

Answer:

Refer to the attachment