If $$\sec \theta + \tan \theta = m (>1)$$, then the value of $$\sin \theta is (0^\circ < \theta < 90^\circ)$$
 $$\sec \theta + \tan \theta = m (>1)$$
let $$\theta = 45\degree$$
$$\surd{2}+ 1 = m$$
$$m^2 = 3+2\surd{2}$$
$$m^2 - 1 = 3+2\surd{2} -1$$ = $$2+2\surd{2}$$
$$m^2 + 1 = 3+2\surd{2} +1$$ = $$2+2\surd{2}$$
$$\frac{m^2 - 1}{m^2 + 1} = \frac{2+2\surd{2}}{2+2\surd{2}}$$ = $$\frac{1}{\surd{2}} = sin\theta$$Â
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