Question

1 - sin^2 y /( 1 + cosy) + ( 1 + cosy) / siny - siny / ( 1- cos y)


Evalute this ​

Answer 1

Given :-

• $\sf\:1-\dfrac{\sin^2y}{1+\cos\:y}+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

To Find :-

• Value of $\sf\:1-\dfrac{\sin^2y}{1+\cos\:y}+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

Formula to be used :-

• $\sf\sin^2x+\cos^2y=1$

• $\sf(a+b)(a-b)=a^2-b^2$

Solution :-

Given that,

$\sf\:1-\dfrac{\sin^2y}{1+\cos\:y}+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

Let's solve this problem.

$\sf\:1-\dfrac{\sin^2y}{1+cos\:y}+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-cos\:y}$

$\sf\:=1-\dfrac{1-\cos^2y}{1+\cos\:y}+\dfrac{1+\cos\:y}{sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

$\sf=1-\dfrac{(1+\cos\:y)(1-\cos\:y)}{1+\cos\:y}+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

$\sf=1-(1-\cos\:y)+\dfrac{1+cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-\cos\:y}$

$\sf=\cos\:y+\dfrac{1+\cos\:y}{\sin\:y}-\dfrac{\sin\:y}{1-cos\:y}$

$\sf=\cos\:y+\dfrac{(1+\cos\:y)(1-\cos\:y)-\sin^2y}{(\sin\:y)(1-cos\:y)}$

$\sf=\cos\:y+\dfrac{(1-\cos^2y-\sin^2y)}{(\sin\:y)(1-cos\:y)}$

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

$\sf=\cos\:y+\dfrac{(1-1)}{(\sin\:y)(1-cos\:y)}$

$\sf=\cos\:y$

Hence, cos y is the required answer.

Answer 2

Answer:

this answer is right she has given