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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
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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?

Answer & Solution
Correct Answer: Option C
$$\eqalign{ & {\text{tan}}\alpha = 2\left( {{\text{given}}} \right) \cr & \therefore \frac{{{\text{cose}}{{\text{c}}^2}\alpha - {\text{se}}{{\text{c}}^2}\alpha }}{{{\text{cose}}{{\text{c}}^2}\alpha + se{c^2}\alpha }} \cr} $$
(Divide by coses2α both in N and D)
$$\eqalign{ & = \frac{{1 - {\text{ta}}{{\text{n}}^2}\alpha }}{{1 + {\text{ta}}{{\text{n}}^2}\alpha }} \cr & = \frac{{1 - {{\left( 2 \right)}^2}}}{{1 + {{\left( 2 \right)}^2}}} \cr & = - \frac{3}{5} \cr} $$
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