Which of the following is equal to $$\left[ {\frac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}} \right]?$$
A. $$\frac{{1 + \sin \theta }}{{\cos \theta }}$$
B. $$\frac{{1 + \tan \theta }}{{\cot \theta }}$$
C. $$\frac{{1 + \cos \theta }}{{\sin \theta }}$$
D. $$\frac{{1 + \cot \theta }}{{\tan \theta }}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \left[ {\frac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec \theta + 1}}} \right] \cr & = \left[ {\frac{{\tan \theta + \sec \theta - \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{\left( {\tan \theta - \sec \theta + 1} \right)}}} \right] \cr & = \frac{{\left( {\sec \theta + \tan \theta } \right)\left[ {1 - \sec \theta + \tan \theta } \right]}}{{\left( {\tan \theta - \sec \theta + 1} \right)}} \cr & = \sec \theta + \tan \theta \cr & = \frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }} \cr & = \frac{{1 + \sin \theta }}{{\cos \theta }} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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