Question : The value of the expression $\left (1+\sec22^{\circ} +\cot68^{\circ} \right )\left (1-cosec\;22 ^{\circ} +\tan68^{\circ} \right )$ is:
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: 2
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Correct Answer: 2
Solution :
Given, $\left (1+\sec22^{\circ} +\cot68^{\circ} \right)\left (1-cosec\;22 ^{\circ} +\tan68^{\circ} \right )$
⇒ $\left (1+\sec22^{\circ} +\tan22^{\circ} \right)\left (1-cosec\;22 ^{\circ} +\cot22^{\circ} \right)$
⇒ $\frac{(\cos22^{\circ}+1 +\sin22^{\circ})}{\cos22^{\circ}}\times \frac{(\sin22^{\circ}-1 +\cos22^{\circ})}{\sin22^{\circ}}$
⇒ $\frac{((\cos22^{\circ}+\sin22^{\circ})^{2}-1}{\cos22^{\circ}\sin22^{\circ}}$
⇒ $\frac{(\cos22^{\circ})^{2}+(\sin22^{\circ})^{2}+2\cos22^{\circ}\sin22^{\circ}-1}{\cos22^{\circ}\sin22^{\circ}}$
Since, $(\cos22^{\circ})^{2}+(\sin22^{\circ})^{2}=1$,
$\left (1+\sec22^{\circ} +\cot68^{\circ} \right )\left (1-cosec\;22 ^{\circ} +\tan68^{\circ} \right )=\frac{2\cos22^{\circ}\sin22^{\circ}}{\cos22^{\circ}\sin22^{\circ}} = 2$.
Hence, the correct answer is 2.
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