अगर $\sec \theta+\tan \theta=2+\sqrt{5}$, तो $\sin \theta$ का मान है $\left(0^{\circ} \leq \theta \leq 90^{ \circ}\दाएं)$

$\sec \theta+\tan \theta=2+\sqrt{5}$

$\because \quad \sec ^{2} \theta-\tan ^{2} \theta=1$

$\Rightarrow \quad(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1$

$\Rightarrow \quad \sec \theta-\tan \theta=\frac{1}{\sqrt{5}+2}$

$=\frac{1}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}$

$=\sqrt{5}-2$

$\therefore \quad \sec \theta+\tan \theta+\sec \theta-\tan \theta=2+\sqrt{5}+\sqrt{5}-2$

$\Rightarrow 2 \sec \theta=2 \sqrt{5}$

$\Rightarrow \sec \theta=\sqrt{5}$

फिर से,

$\sec \theta+\tan \theta-(\sec \theta-\tan \theta)$

$=2+\sqrt{5}-\sqrt{5}+2$

$\Rightarrow 2 \tan \theta=4 \Rightarrow \tan \theta=2 \ldots$ (ii)

$\therefore \quad \sin \theta=\frac{\tan \theta}{\sec \theta}=\frac{2}{\sqrt{5}}$