Prove that:$ \tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2 $


To do:

We have to prove that \( \tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2 \).

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,

$\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}=\frac{\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}+\frac{\cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A}}$

$=\frac{\sin ^{4} \mathrm{~A}+\cos ^{4} \mathrm{~A}}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

$=\frac{\left(\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}\right)^{2}-2 \sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$

$=\frac{(1)^{2}-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{1-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{1}{\sin ^{2} A \cos ^{2} A}-\frac{2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$

$=\sec ^{2} A \operatorname{cosec}^{2} A-2$

Hence proved.     

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Updated on: 10-Oct-2022

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