Question

Verify the identity.

cot^2(x)/(cscx+1)=(1-sinx)/sinx

Answer 1

I'm sorry, I'm in 9th grade, I don't know calculus, I know everything else though

Answer 2

The only thing about calculus I know is the chain method lol

Answer 3

thankyou!

Answer 4

@satellite73 help me:(

Answer 5

You don't need calculus, only trig definitions and identities:\[\cot x = \frac{ \cos x }{ \sin x }\]\[\csc x = \frac{ 1 }{ \sin x }\]and some algebra skills.

My advice: use these formulas for cot x and csc x and try to get closer to the result.

Answer 6

So to get you started with the left-hand side of the formula:\[\frac{ \cot^2x }{ \csc x +1 }=\frac{ \frac{ \cos^2x }{ \sin^2x } }{ \frac{ 1 }{ \sin x }+1 }\]Now, first write the denominator as one fraction and then simplify.

Answer 7

Next steps:\[\frac{ \frac{ \cos^2x }{ \sin^2x } }{ \frac{ 1 }{ \sin x }+1 }=\frac{\frac{ \cos^2x }{ \sin^2x } }{ \frac{ 1+\sin x }{ \sin x } }=\frac{ \cos^2x }{ \sin^2x } \cdot \frac{ \sin x }{ 1+\sin x }=\frac{ \cos^2x }{ \sin x(1+\sin x) }\]If you use the rule cos²x+sin²x=1, you can rewrite the numerator as 1- sin²x.
This could then be factored as (1-sin x)(1+sin x).
Then simplify to arrive at your final destination ;)