| | Question 46

 \sqrt{\frac{1+sin\theta }{1-sin\theta }}\:+\:\sqrt{\frac{1-sin\theta \:}{1+sin\theta \:}} is equal to
(A) 2 secθ
(B) 2 cosθ
(C) 2 sinθ
(D) 2 tanθ

Answer: (A)

Explanation: It can be solved by rationalization
 \frac{\left(\sqrt{1+sin\theta \:}\right)^2\:+\:\left(\sqrt{1-sin\theta \:\:}\right)^2}{\sqrt{1-sin^2\theta \:\:}}
Use the property:-
(1-sin2θ = cos2θ)
 =\frac{1\:+sin\theta \:+\:1\:-sin\theta }{cos\theta }
= 2/cosθ = 2 secθ

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