Question

finf valve of this sum​

Answer 1

$\rm \longrightarrow \sqrt{ \dfrac{1 + sin \theta}{1 - sin \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{1 + sin \theta}{1 - sin \theta} \times\dfrac{1 + sin \theta}{1 + sin \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{ {(1 + sin \theta)}^{2} }{1 - {sin}^{2} \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{ {(1 + sin \theta)}^{2} }{ {cos}^{2} \theta} }$

$\rm \longrightarrow { \dfrac{ {1 + sin \theta}}{ {cos} \theta} }$

$\rm \longrightarrow { \dfrac{ {1 }}{ {cos} \theta} } + { \dfrac{ { sin \theta}}{ {cos} \theta} }$

$\rm \longrightarrow {{sec \: \theta} } + {{tan \: \theta} }$

Therefore, option (A) is correct.

Additional Information :-

$\sf Trigonometric\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$

Answer 2

$\rm \longrightarrow \sqrt{ \dfrac{1 + sin \theta}{1 - sin \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{1 + sin \theta}{1 - sin \theta} \times\dfrac{1 + sin \theta}{1 + sin \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{ {(1 + sin \theta)}^{2} }{1 - {sin}^{2} \theta} }$

$\rm \longrightarrow \sqrt{ \dfrac{ {(1 + sin \theta)}^{2} }{ {cos}^{2} \theta} }$

$\rm \longrightarrow { \dfrac{ {1 + sin \theta}}{ {cos} \theta} }$

$\rm \longrightarrow { \dfrac{ {1 }}{ {cos} \theta} } + { \dfrac{ { sin \theta}}{ {cos} \theta} }$

$\rm \longrightarrow {{sec \: \theta} } + {{tan \: \theta} }$

• Therefore, option (A) is correct.