Prove that (tan x)/(sin^2 x)=±√(1+tan^2 x)/(1-cos^2 x)

Question

Please help me prove this, your help is greatly appreciated. Thank you.

Answer 1

Hi Zandra:

Here's how to tackle this kind of problem. Choose one side of the identity, usually the more complicated one, and manipulate the trig functions until you get to the function(s) on the other side of the equals sign.

You need all the standard trig identities: sin² x + cos² x = 1, Tan x = sin x / cos x, etc.

Alternatively, prove that both sides can be simplified to the same expression.

Let's start with the RHS: √(1 + tan² x) / (1 - cos² x) = √((sec² x) / sin² x [1-cos² x = sin² x

= sec x / sin x [taking square roots

= 1/(cos x sin x)

Now the LHS = tan x / sin² x

= (sin x /cos x) / sin² x

= 1 / (sin x cos x)

= RHS

Done!

Answer 2

sqrt[( 1 + tan^2)/(1-cos^2)] =

sqrt[ sec^2 / sin^2]= <---- 1 + tan^2 = sec^2 and 1-cos^2 = sin^2

sec/sin =

1/ (sin * cos) = <--- sec = 1/cos

1/(sin * cos) * 1 =

1/(sin * cos) * (sin / sin) = <--- 1 = sin/sin

sin/ (cos *sin^2) =

(sin/ cos) * (1 / sin^2) =

tan / sin^2