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Prove the following:$ \frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=0 $
To do:
We have to prove that \( \frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=0 \).
Solution:
We know that,
$tan\ (90^{\circ}- \theta) = cot\ \theta$
$sin\ \theta=\frac{1}{cosec\ \theta}$
$\cot \theta=\frac{\cos\ \theta}{\sin\ \theta}$
Therefore,
$\frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=\frac{\cot A \times \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A$
$=\cot^2 A \times \frac{1}{\operatorname {cosec}^{2} A}-\cos ^{2} A$
$=\frac{cos^{2} A}{sin^{2} A}\times sin^{2} A-cos^{2} A$
$=cos^2 A-cos^2 A$
$=0$
Hence proved.
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