Question

can you please solve this question with explanation . please no spammers ​

Answer 1

Question :

Find the value of expression

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

Formulas used :

$\sf1)\sin^2x+\cos^2y=1$

$2)\sf(a+b)(a-b)=a^2-b^2$

Solution :

We have to find the value of

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

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

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

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

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

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

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

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

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

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

$\sf=\cos\:y$

Correct option d) cos y

Answer 2

Answer:

Refers to the attachment....