1. ## Trig proving

Could someone help me prove this identity:

$\frac{sin\theta-cos\theta}{cosec\theta-sin\theta} = tan^2\theta + tan\theta$

2. ## Re: Trig proving

LHS is not equivalent to the RHS, can you check the question?

or are you looking for solutions instead?

3. ## Re: Trig proving

Originally Posted by pilota45
Could someone help me prove this identity:

$\frac{sin\theta-cos\theta}{cosec\theta-sin\theta} = tan^2\theta + tan\theta$
$LHS = \frac {sin\theta - cos\theta}{\frac{1}{sin\theta} - sin\theta} =\frac {(sin\theta - cos\theta)sin\theta}{1 - sin^2 \theta} = \frac {sin^2 \theta - sin\theta cos \theta}{cos^2 \theta} \\ \\ = \frac {sin^2 \theta}{cos^2 \theta} - \frac {sin\theta cos\theta}{cos^2 \theta} = tan^2 \theta - tan\theta \neq RHS$

4. ## Re: Trig proving

Originally Posted by jathu123
LHS is not equivalent to the RHS, can you check the question?

or are you looking for solutions instead?
Originally Posted by Drongoski
$LHS = \frac {sin\theta - cos\theta}{\frac{1}{sin\theta} - sin\theta} =\frac {(sin\theta - cos\theta)sin\theta}{1 - sin^2 \theta} = \frac {sin^2 \theta - sin\theta cos \theta}{cos^2 \theta} \\ \\ = \frac {sin^2 \theta}{cos^2 \theta} - \frac {sin\theta cos\theta}{cos^2 \theta} = tan^2 \theta - tan\theta \neq RHS$
Sorry I've written the RHS wrong. It's meant to be $tan^2 \theta - tan\theta$. Thanks for your help

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