Question : If $\operatorname{cosec} 39°=x$, then the value of $\frac{1}{\operatorname{cosec}^{2}51°}+\sin^{2}39°+\tan^{2}51°-\frac{1}{\sin^{2}51°\sec^{2}39°}$ is:
Option 1: $\sqrt{x^{2}-1}$
Option 2: $\sqrt{1-x^{2}}$
Option 3: $x^{2}-1$
Option 4: $1-x^{2}$
Correct Answer: $x^{2}-1$
Solution :
Given: $\frac{1}{\operatorname{cosec}^{2}51°}+\sin^{2}39°+\tan^{2}51°-\frac{1}{\sin^{2}51°\sec^{2}39°}$
= ${\sin^{2}51°}+\sin^2(90°-51°)+\tan^{2}51°-\frac{\cos^{2}39°}{\sin^{2}(90°-39°)}$
= ${\sin^{2}51°}+\cos^251°+\tan^{2}51°-\frac{\cos^{2}39°}{\cos^{2}39°}$
= $1+\tan^251°-1$
= $1+\cot^239°-1$ $[\because \tan^251°=\cot^239°]$
= $\operatorname{cosec}^239°-1 $
= $x^{2}-1$
Hence, the correct answer is $x^{2}-1$.
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