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.