Question : $\frac{1}{1+\cos(90^{\circ}- \theta)}$ + $\frac{1}{1-\cos(90^{\circ}- \theta)}$ =?
Option 1: $2\sec^2 \theta$
Option 2: $1$
Option 3: $2$
Option 4: $2 \tan^2 \theta$
Correct Answer: $2\sec^2 \theta$
Solution :
$\frac{1}{1+\cos(90^{\circ}- \theta)}$ + $\frac{1}{1-\cos(90^{\circ}- \theta)}$
= $\frac{1}{1+\sin \theta}$ + $\frac{1}{1-\sin \theta}$
= $\frac{(1-\sin \theta)+(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
= $\frac{2}{1-\sin ^2 \theta}$
= $\frac{2}{\cos ^2 \theta}$
= $2\sec^2 \theta$
Hence, the correct answer is $2\sec^2 \theta$.
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