Question

1+costheta + sintheta /1+ vostheta -sintheta = 1+ sintheta /costheta ​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}=\dfrac{1+sin\theta}{cos\theta}$

Solution :-

Taking L.H.S :-

$=\dfrac{1+cos\theta+sin\theta}{1+cos\theta-sin\theta}$

Multiplying and dividing '1+cosθ+sinθ' :-

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

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

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

Expanding the terms :-

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

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

$[\ sin^2\theta=1-cos^2\theta]$

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

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

$=\dfrac{2+2cos\theta+2sin\theta(1+cos\theta)}{2cos\theta(cos\theta+1)}$

$=\dfrac{2(1+cos\theta)+2sin\theta(1+cos\theta)}{2cos\theta(cos\theta+1)}$

Taking common :-

$=\dfrac{2(1+cos\theta)(1+sin\theta)}{2(cos\theta+1)(cos\theta)}$

$=\dfrac{1+sin\theta}{cos\theta}$

= R.H.S

Hence Proved.

Formula Used :-

$1) sin^2\theta=1-cos^2\theta$