Prove that sin^2x(1+cot^2x)=1?

Question

Sin^2x+(sin^2*cot^2x)=1

Thats all I have!!!

there is no typo in either one.

Answer 1

Hi,

sin² x(1 + cot² x)= 1

One of the Pythagorean identities is 1 + cot² x = csc² x, so replace the parentheses with csc² x.

sin² x(1 + cot² x)= 1
sin² x(csc² x)= 1

Since csc x = 1/sin x, then csc² x = 1/sin² x

sin² x(csc² x)= 1
sin² x(1/sin² x)= 1

sin² x cancels out and

1 = 1

I hope that helps!! :-)

Answer 2

Ummm... I think you made a typo when you rewrote it, but the thing in your original question is actually very easy to show:

sin²x (1 + cot²x) = 1
==> rewrite cot²x in terms of sine and cosine
sin²x (1 + cos²x / sin²x) = 1
==> distribute the sin²x
sin²x + (sin²x)(cos²x / sin²x) = 1
==> cancel the sin²x in numerator and denominator in 2nd term
sin²x + cos²x = 1
==> this is a basic identity, so we're done
1 = 1.

Q.E.D.

Answer 3

cot^2 x=cos^2 x/sin^2 x. So if you distribute the multiplication, you get sin^2 x+ cos^2 x on the left, which is of course equal to 1.

Answer 4

sin^2x + (sin^2*cot^2x) = 1
Replace cotx with its definition:
sin^2x + sin^2*(cos^2x/sin^2) = 1
sin^2x + cos^2x = 1