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