Question:
What is the value of $\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta} ?$
Solution:
We have,
$\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta}$
$=\sin ^{2} \theta+\left(\frac{1}{\sec \theta}\right)^{2}$
$=\sin ^{2} \theta+\cos ^{2} \theta$
$=1$
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