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We know that tanФ=sinФ/cosФ and cotФ=1/tanФ.
So we need to prove that (tanФ-cotФ)/sinФcosФ=tan²Ф-cot²Ф
                       we can see that (tanФ-cotФ)/sinФcosФ=(tanФ-cotФ)(tanФ+cotФ)
So what we really need to prove is that 1/sinФcosФ=(tanФ+cotФ)
Now we know that (tanФ+cotФ)=sinФ/cosФ + cosФ/sinФ
[We know the fact that sin²Ф+cos²Ф=1] hence we have proved that 1/sinФcosФ=(tanФ+cotФ).Thus the problem is solved.
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little bit
What havent u understood?
now i got it
thnkz yaar
:D no prob

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 \frac{tan\ \alpha - cot\ \alpha }{sin\ \alpha \ cos\ \alpha } \\ \\ 
\frac{\frac{sin\ \alpha }{cos\ \alpha } - \frac{cos\ \alpha }{sin\ \alpha } }{sin\ \alpha \ cos\ \alpha } \\ \\ 
\frac{sin^{2} \alpha - cos^{2} \alpha }{sin^{2} \alpha \ cos^{2} \alpha } \\ \\ 
\frac{1}{cos^{2} \alpha } - \frac{1}{sin^{2} \alpha } \\ \\ 
sec^{2}\ \alpha - cosec^{2}\ \alpha \\ \\ 
R H S : \ \ tan^{2}\ \alpha - cot^{2}\ \alpha = (sec^{2}\ \alpha - 1) - (cosec^{2}\ \alpha - 1) \\ \\ 
So\ the\ answer. \\
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