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Prove that:$ \cot ^{2} A \operatorname{cosec}^{2} B-\cot ^{2} B \operatorname{cosec}^{2} A=\cot ^{2} A-\cot ^{2} B $
To do:
We have to prove that \( \cot ^{2} A \operatorname{cosec}^{2} B-\cot ^{2} B \operatorname{cosec}^{2} A=\cot ^{2} A-\cot ^{2} B \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
Let us consider LHS,
$\cot ^{2} \mathrm{~A} \operatorname{cosec}^{2} \mathrm{~B}-\cot ^{2} \mathrm{~B} \operatorname{cosec}^{2} \mathrm{~A}=\frac{\cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A}} \times \frac{1}{\sin ^{2} \mathrm{~B}}-\frac{\cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~B}} \times \frac{1}{\sin ^{2} \mathrm{~A}}$
$=\frac{\cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}}-\frac{\cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}}$
$=\frac{\cos ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}}$
Let us consider RHS,
$\cot ^{2} \mathrm{~A}-\cot ^{2} \mathrm{~B}=\frac{\cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A}}-\frac{\cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~B}}$
$=\frac{\cos ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}-\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}}$
$=\frac{\cos ^{2} \mathrm{~A}\left(1-\cos ^{2} \mathrm{~B}\right)-\left(1-\cos ^{2} \mathrm{~A}\right) \cos ^{2} \mathrm{~B}}{\sin ^{2} \mathrm{~A} \sin ^{2} \mathrm{~B}}$
$=\frac{\cos ^{2} A-\cos ^{2} A \cos ^{2} B-\cos ^{2} B+\cos ^{2} A \cos ^{2} B}{\sin ^{2} A \sin ^{2} B}$
$=\frac{\cos ^{2} A-\cos ^{2} B}{\sin ^{2} A \sin ^{2} B}$
Here,
LHS $=$ RHS
Hence proved.
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