Question : If $0^{\circ}< A< 90^{\circ}$, then the value of $\tan ^{2}A+\cot^{2}A-\sec^{2}A \operatorname{cosec}^{2}A$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –2
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Correct Answer: –2
Solution :
Given: $0^{\circ}< A< 90^{\circ}$
$\tan ^{2}A+\cot^{2}A-\sec^{2}A \operatorname{cosec}^2A$
= $\frac{\sin^2 A}{\cos^2 A}+\frac{\cos^2 A}{\sin^2 A}-\frac{1}{\cos^2A\sin^2A}$
= $\frac{\sin^4+\cos^4-1}{\cos^2A\sin^2A}$
= $\frac{(\sin^2A+\cos^2A)^2-2\sin^2\cos^2A-1}{\cos^2A\sin^2A}$
= $\frac{1-2\sin^2A\cos^2A-1}{\cos^2A\sin^2A}$
= $\frac{-2\sin^2A\cos^2A}{\cos^2A\sin^2A}$
= $-2$
Hence, the correct answer is –2.
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