If secθ + tanθ = m(>1), then the value of sinθ is (0° < θ < 90°)

A. $$\frac{{1 - {m^2}}}{{1 + {m^2}}}$$

B. $$\frac{{{m^2} - 1}}{{{m^2} + 1}}$$

C. $$\frac{{{m^2} + 1}}{{{m^2} - 1}}$$

D. $$\frac{{1 + {m^2}}}{{1 - {m^2}}}$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & \sec \theta + \tan \theta = m\,.......(i) \cr & {\text{then, }}\sec \theta - \tan \theta = \frac{1}{m}\,......(ii) \cr & Because{\text{ }}{\sec ^2}\theta - {\text{ta}}{{\text{n}}^2}\theta = 1 \cr & {\text{From equation (i)}} - {\text{(ii)}} \cr & 2\tan \theta = m - \frac{1}{m} \cr} $$

$$\eqalign{ & {\text{tan}}\theta = \frac{{{m^2} - 1}}{{2m}} \cr & \sin \theta = \frac{{{m^2} - 1}}{{{m^2} + 1}} \cr} $$