Question : $2\ \text{cosec}^{2}\: 23°\cot^{2}67°-\sin^{2}23°-\sin^{2}67°-\cot^{2}67°$ is equal to:
Option 1: $1$
Option 2: $\sec^{2}23°$
Option 3: $\tan^{2}23°$
Option 4: $0$
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Correct Answer: $\sec^{2}23°$
Solution :
$2\ \text{cosec}^{2}\: 23°\cot^{2}67°-\sin^{2}23°-\sin^{2}67°-\cot^{2}67°$
$=2\: \frac{1}{\sin^{2}\: 23°}\tan^{2}23°-\sin^{2}23°-\cos^{2}23°-\tan^{2}23°$
$=2\: \frac{1}{\cos^{2}23°}-(\sin^{2}23°+\cos^{2}23°)-\frac{\sin^{2}23°}{\cos^{2}23°}$
$=2\: \frac{1}{\cos^{2}23°}-1-\frac{\sin^{2}23°}{\cos^{2}23°}$
$=\frac{2-\cos^2 23° - \sin^223°}{\cos^2 23°}$
$=\frac{2-1}{\cos^2 23°}$
$=\sec^2 23°$
Hence, the correct answer is $\sec^2 23°$.
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