Question : If $\alpha +\beta =90^{0}$, then value of $\left (1-\sin^{2}\alpha \right)\left (1-\cos^{2}\alpha \right)\times \left (1+\cot^{2}\beta \right)\left (1+\tan^{2}\beta \right)$ is:
Option 1: 1
Option 2: –1
Option 3: 0
Option 4: 2
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Correct Answer: 1
Solution :
Given: $\alpha +\beta = 90° ⇒ \alpha = 90° - \beta$
Using identities: $\cos^2\alpha + \sin^2\alpha = 1$ , $\cot^2\alpha +1 = \text{cosec}^2\alpha$ and $\tan^2\alpha +1 = \sec^2\alpha$, we get:
$\left (1-\sin^{2}\alpha \right)\left (1-\cos^{2}\alpha \right)\times \left (1+\cot^{2}\beta \right)\left (1+\tan^{2}\beta \right)$
$=\cos^2\alpha \sin^2\alpha \times \text{cosec}^2\beta \sec^2\beta$
$=\cos^2(90-\beta) \sin^2 (90-\beta) \times \text{cosec}^2\beta \sec^2\beta$
$=\sin^2\beta \cos^2 \beta \times \text{cosec}^2\beta \sec^2\beta$
$=1$
Hence, the correct answer is 1.
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