\(\frac{1+Sin\theta}{\sqrt{1-Sin^2\theta}}\) =A) Sec θ + Tan θ B)

1.	\(\frac{1+Sin\theta}{\sqrt{1-Sin^2\theta}}\) =A) Sec θ + Tan θ B) Sec θ – Tan θC) Sec2 θ + Tan2 θD) Sec2 θ – Tan2 θ
Answer» Correct option is: A) Sec θ + Tan θ \(\frac{1+Sin \theta}{\sqrt{1-Sin^2 \theta}} =\) \(\frac{1+Sin \theta}{ \sqrt{cos^2 \theta}} = \frac {1+sin\, \theta}{cos\, \theta}\) = \(\frac 1{cos\, \theta} + \frac {sin \, \theta }{cos \ \theta}\) = sec \(\theta\) + tan \(\theta\) Correct option is: A) Sec θ + Tan θ

\(\frac{1+Sin\theta}{\sqrt{1-Sin^2\theta}}\) =A) Sec θ + Tan θ B) Sec θ – Tan θC) Sec2 θ + Tan2 θD) Sec2 θ – Tan2 θ

Answer»

Correct option is: A) Sec θ + Tan θ

\(\frac{1+Sin \theta}{\sqrt{1-Sin^2 \theta}} =\) \(\frac{1+Sin \theta}{ \sqrt{cos^2 \theta}} = \frac {1+sin\, \theta}{cos\, \theta}\)

= \(\frac 1{cos\, \theta} + \frac {sin \, \theta }{cos \ \theta}\)

= sec \(\theta\) + tan \(\theta\)

Correct option is: A) Sec θ + Tan θ