Question

pls answer this ......................

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\sf{Given :\;\;\dfrac{2sin\theta}{1 + cos\theta + sin\theta} = x}\\\\\\\sf{Multiplying\;and\;Dividing\;L.H.S\;with\;(1 + sin\theta - cos\theta),\;We\;get :}\\\\\\\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{(1 + sin\theta + cos\theta)(1 + sin\theta - cos\theta)} = x}}\\\\\\\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{(1 + sin\theta)^2 - (cos\theta)^2} = x}}$

$\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{(1 + sin^2\theta + 2sin\theta) - (cos^2\theta)} = x}}\\\\\\\sf{\bigstar\;\;We\;know\;that : \boxed{\sf{1 = cos^2\theta + sin^2\theta}}}\\\\\\\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{(cos^2\theta + sin^2\theta + sin^2\theta + 2sin\theta) - (cos^2\theta)} = x}}\\\\\\\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{2sin^2\theta + 2sin\theta} = x}}$

$\;\sf{\implies \dfrac{2sin\theta(1 + sin\theta - cos\theta)}{2sin\theta(sin\theta + 1)} = x}}\\\\\\\sf{\; \implies \dfrac{1 + sin\theta - cos\theta}{sin\theta + 1} = x}}$