Question : If $\sec\theta+\tan\theta=m\left (> 1 \right)$, then the value of $\sin\theta$ is $\left (0^{\circ} < \theta<90^{\circ} \right)$:
Option 1: $\frac{1-m^{2}}{1+m^{2}}$
Option 2: $\frac{m^{2}-1}{m^{2}+1}$
Option 3: $\frac{m^{2}+1}{m^{2}-1}$
Option 4: $\frac{1+m^{2}}{1-m^{2}}$
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Correct Answer: $\frac{m^{2}-1}{m^{2}+1}$
Solution :
Given: $\sec\theta+\tan\theta=m$
⇒ $1+\sin\theta=m\cos\theta$
Squaring both sides,
$1+2\sin\theta+\sin^{2}\theta=m^{2}\cos^{2}\theta$
⇒ $1+2\sin\theta+\sin^{2}\theta=m^{2}(1-\sin^{2}\theta)$
⇒ $(1+m^{2})\sin^{2}\theta+2\sin\theta+(1-m^{2})=0$
⇒ $\sin\theta = \frac{-2\pm\sqrt{2^2-4(1+m^2)(1-m^2)}}{2(1+m^{2})}$
⇒ $\sin\theta =\frac{-2\pm 2m^2}{2(1+m^{2})}$
⇒ $\sin\theta =\frac{m^2-1^2}{(1+m^{2})}$ and $\sin\theta =-1$
⇒ $\sin\theta=\frac{m^{2}-1}{m^{2}+1},-1$
Hence, the correct answer is $\frac{m^{2}-1}{m^{2}+1}$.
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