If tanθ + secθ = $$\frac{{x - 2}}{{x + 2}},$$ then what is the value of cosθ?
A. $$\frac{{{x^2} - 1}}{{{x^2} + 1}}$$
B. $$\frac{{2{x^2} - 4}}{{2{x^2} + 4}}$$
C. $$\frac{{{x^2} - 4}}{{{x^2} + 4}}$$
D. $$\frac{{{x^2} - 2}}{{{x^2} + 2}}$$
Answer: Option C
Solution (By Examveda Team)
$$\eqalign{ & \tan \theta + \sec \theta = \frac{{x - 2}}{{x + 2}}........\left( {\text{i}} \right) \cr & \frac{{\left( {\sec \theta + \tan \theta } \right)}}{1} \times \frac{{\left( {\sec \theta - \tan \theta } \right)}}{{\left( {\sec \theta - \tan \theta } \right)}} = \frac{{x - 2}}{{x + 2}} \cr & \frac{1}{{\sec \theta - \tan \theta }} = \frac{{x - 2}}{{x + 2}} \cr & \left( {\because \,{{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right) \cr & \sec \theta - \tan \theta = \frac{{x + 2}}{{x - 2}}........\left( {{\text{ii}}} \right) \cr & {\text{By equation }}\left( {{\text{ii}}} \right) + \left( {\text{i}} \right){\text{ we get}} \cr & 2\sec \theta = \frac{{x + 2}}{{x - 2}} + \frac{{x - 2}}{{x + 2}} \cr & 2\sec \theta = \frac{{{{\left( {x + 2} \right)}^2} + {{\left( {x - 2} \right)}^2}}}{{\left( {{x^2} - {2^2}} \right)}} \cr & 2\sec \theta = \frac{{\left[ {{x^2} + 4 + 2x + {x^2} + 4 - 2x} \right]}}{{\left( {{x^2} - 4} \right)}} \cr & 2\sec \theta = \frac{{2\left( {{x^2} + 4} \right)}}{{\left( {{x^2} - 4} \right)}} \cr & \sec \theta = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \frac{1}{{\cos \theta }} = \frac{{{x^2} + 4}}{{{x^2} - 4}} \cr & \cos \theta = \frac{{{x^2} - 4}}{{{x^2} + 4}} \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
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