If cos53° = $$\frac{x}{y},$$ then sec53° + cot37° is equal to:
A. $$\frac{{x + \sqrt {{y^2} - {x^2}} }}{y}$$
B. $$\frac{{x + \sqrt {{y^2} - {x^2}} }}{x}$$
C. $$\frac{{y + \sqrt {{y^2} - {x^2}} }}{x}$$
D. $$\frac{{y + \sqrt {{y^2} - {x^2}} }}{y}$$
Answer: Option C
Solution (By Examveda Team)
$$\eqalign{ & \cot {53^ \circ } = \frac{x}{y} = \frac{B}{H} \cr & {P^2} = {H^2} - {B^2} \cr & P = \sqrt {{y^2} - {x^2}} \cr & \sec {53^ \circ } + \cot {37^ \circ } \cr & = \sec {53^ \circ } + \tan {53^ \circ } \cr & = \frac{H}{B} + \frac{P}{B} \cr & = \frac{{H + P}}{B} \cr & = \frac{{y + \sqrt {{y^2} - {x^2}} }}{x} \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
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