$$\sqrt {\frac{{1 + {\text{sin }}\theta }}{{1 - {\text{sin }}\theta }}} $$ + $$\sqrt {\frac{{1 - {\text{sin }}\theta }}{{1 + {\text{sin }}\theta }}} $$ is equal to?
A. 2cosθ
B. 2sinθ
C. 2cotθ
D. 2secθ
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \sqrt {\frac{{1 + {\text{sin }}\theta }}{{1 - {\text{sin }}\theta }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}\theta }}{{1 + {\text{sin }}\theta }}} \cr & = \frac{{{{\left( {\sqrt {1 + {\text{sin }}\theta } } \right)}^2} + {{\left( {\sqrt {1 - {\text{sin }}\theta } } \right)}^2}}}{{\sqrt {1 - {{\sin }^2}\theta } }} \cr & = \frac{{1 + {\text{sin }}\theta + {\text{1}} - {\text{sin }}\theta }}{{{\text{cos}}\theta }} \cr & = \frac{2}{{{\text{cos}}\theta }} \cr & = 2{\text{ sec}}\theta \cr} $$Alternate shortcut method :
$$\eqalign{ & {\text{Put }}\theta = {30^ \circ } \cr & \sqrt {\frac{{1 + {\text{sin }}30^ \circ }}{{1 - {\text{sin }}30^ \circ }}} {\text{ + }}\sqrt {\frac{{1 - {\text{sin }}30^ \circ }}{{1 + {\text{sin }}30^ \circ }}} \cr & \Rightarrow \sqrt {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} + \sqrt {\frac{{1 - \frac{1}{2}}}{{1 + \frac{1}{2}}}} \cr & \Rightarrow \sqrt {\frac{3}{1}} + \sqrt {\frac{1}{3}} \cr & \Rightarrow \frac{4}{{\sqrt 3 }} \cr} $$
Now check with option by puting θ = 30°
$$\eqalign{ & {\text{ = 2 sec 3}}{0^ \circ } \cr & = \frac{{2 \times 2}}{{\sqrt 3 }} \cr & = \frac{4}{{\sqrt 3 }} \cr} $$
Related Questions on Trigonometry
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B. x > y
C. x = y
D. x < y
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