Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities2 Marks
Question
Prove that $\frac{(1+\sin\theta)}{(1-\sin\theta)}=(\sec\theta+\tan\theta)^2.$
✓
Answer
$\frac{(1+\sin\theta)}{(1-\sin\theta)}=(\sec\theta+\tan\theta)^2$$\text{LHS}=\frac{(1+\sin\theta)}{(1-\sin\theta)}$
Multiplying the numerator and denominator by $(1+\sin\theta),$ we get:
$\frac{(1+\sin\theta)^2}{1-\sin^2\theta}$
$=\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\sec^2\theta+2\times\frac{\sin\theta}{\cos\theta}\times\sec\theta+\tan^2\theta$
$=\sec^2\theta+2\times\tan\theta\times\sec\theta+\tan^2\theta$
$=(\sec\theta+\tan\theta)^2$
$=\text{RHS}$
Hence proved.
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