# Verifying a trig identity

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• August 9th 2013, 11:45 AM
curt26
Verifying a trig identity
I have

1/sin^2(t) + 1/cos^2(t) = 1/sin^2(t)-sin^4(t)

I started on the right side of the identity and got to this stage:

1/sin^2(t) + 1/1-sin^2(t)

On the left side of the identity I got to this stage:

1/sin^2(t)(1-sin^2(t))

I know I need to go farther on the right side of the identity but I am stuck at this point. Any help would be great thanks!
• August 9th 2013, 12:19 PM
adkinsjr
Re: Verifying a trig identity
There's some ambiguity in the way you typed the right side of the equation, but the left reduces down to this:

$\frac{1}{sin^2(t)}+\frac{1}{1-sin^2(t)}=\frac{1-sin^2(t)+sin^2(t)}{sin^2(t)[1-sin^2(t)]}=\frac{1}{sin^2(t)-sin^4(t)}$
• August 13th 2013, 08:19 PM
ibdutt
Re: Verifying a trig identity
alternatively we can also do like this
LHS = 1/sin^2(t) + 1/cos^2(t) = [ sin^2(t) + cos^2(t) ]/ [ sin^2(t) * cos^2(t)] = 1 / [ sin^2(t) * ( 1 - sin ^2(t) )]= RHS