If $${\text{tan }}\alpha = 2,$$ then the value of $$\frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }}$$ is?
A. $$ - \frac{5}{9}$$
B. $$\frac{3}{5}$$
C. $$ - \frac{3}{5}$$
D. $$\frac{{17}}{5}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\text{tan}}\alpha = 2\left( {{\text{given}}} \right) \cr & \therefore \frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }} \cr} $$(Divide by coses2α both in N and D)
$$\eqalign{ & = \frac{{1 - {\text{ta}}{{\text{n}}^2}\alpha }}{{1 + {\text{ta}}{{\text{n}}^2}\alpha }} \cr & = \frac{{1 - {{\left( 2 \right)}^2}}}{{1 + {{\left( 2 \right)}^2}}} \cr & = - \frac{3}{5} \cr} $$
Related Questions on Trigonometry
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B. x > y
C. x = y
D. x < y
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