If sin21° = $$\frac{x}{y}{\text{,}}$$ then sec21° - sin69° is equal to?
A. $$\frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }}$$
B. $$\frac{{{y^2}}}{{x\sqrt {{y^2} - {x^2}} }}$$
C. $$\frac{{{x^2}}}{{y\sqrt {{x^2} - {y^2}} }}$$
D. $$\frac{{{y^2}}}{{x\sqrt {{x^2} - {y^2}} }}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$
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