Question : The value of $\frac{3\left(\operatorname{cosec}^2 26^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\sec ^2 48^{\circ}\right)}{\cot \left(22^{\circ}-\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\tan ^2\left(28^{\circ}-\theta\right)}$ is:
Option 1: 3
Option 2: 4
Option 3: –1
Option 4: –2
Correct Answer: –2
Solution :
$\frac{3\left(\operatorname{cosec}^2 26^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\sec ^2 48^{\circ}\right)}{\cot \left(22^{\circ}-\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\tan ^2\left(28^{\circ}-\theta\right)}$
$= \frac{3\left(\operatorname{sec}^2 64^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\operatorname{cosec} ^2 42^{\circ}\right)}{\tan \left(68^{\circ}+\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\cot ^2\left(62^{\circ}+\theta\right)}$
$= \frac{3\left(1\right)+\left(\cot ^2 42^{\circ}-1-\cot ^2 42^{\circ}\right)}{-(\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\cot ^2\left(62^{\circ}+\theta\right))}$
$= \frac{3-1}{-1}$
$= -2$
Hence, the correct answer is –2.
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