Question

Can someone please help me understand how to verify this problem?

2-cos^2x/sinx = cscx+sinx

Answer 1

Rewrite 2 - cos^2x as 1 + 1 - cos^2x
Then use an identity that involves 1 - cos^2x

Answer 2

Ok I've got it down to: sin^2x * cscx

Now where?? :/

Answer 3

rewrite as (1+(1-cos^2x))/sinx = (1+sin^2x)/sin x= Ans

Answer 4

\(\dfrac{2-\cos^2 x}{\sin x} = \csc x+\sin x\)

\(\dfrac{1 + (1-\cos^2 x)}{\sin x} = \csc x+\sin x\)

Now use the identity

\( \sin^2 x + \cos ^2 x = 1\),

which can be written as

\(\color{red}{1 - \cos^2 x} = \color{green}{\sin^2 x}\)

\(\dfrac{1 + \color{red}{(1-\cos^2 x)}}{\sin x} = \csc x+\sin x\)

\(\dfrac{1 + \color{green}{\sin^2 x}}{\sin x} = \csc x+\sin x\)

\(\color{blue}{\dfrac{1}{\sin x}} + \color{brown}{\dfrac{\sin^2 x}{\sin x}} = \csc x+\sin x\)

\(\color{blue}{\csc x}+\color{brown}{\sin x} = \csc x+\sin x \)