Question
What is the value of $(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)?$
✓
Answer
$(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)$
$=(1+\tan^2\theta)\{(1-\sin\theta)(1+\sin\theta)\}$
$=(1+\tan^2\theta)(1-\sin^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow\ \sec^2\theta=1+\tan^2\theta,$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\sin^2\theta$
Therefore,
$(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)=\sec^2\theta\times\cos^2\theta$
$=\frac{1}{\cos^2\theta}\times\cos^2\theta$
$=1$
$=(1+\tan^2\theta)\{(1-\sin\theta)(1+\sin\theta)\}$
$=(1+\tan^2\theta)(1-\sin^2\theta)$
We know that,
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow\ \sec^2\theta=1+\tan^2\theta,$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\sin^2\theta$
Therefore,
$(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)=\sec^2\theta\times\cos^2\theta$
$=\frac{1}{\cos^2\theta}\times\cos^2\theta$
$=1$
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