Let 0° < θ < 90°, (1 + cot2θ)(1 + tan2θ) × (sinθ - cosecθ)(cosθ - secθ) is equal to:
A. sinθcosθ
B. secθ + cosecθ
C. sinθ + cosθ
D. secθcosecθ
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \left( {1 + {{\cot }^2}\theta } \right)\left( {1 + {{\tan }^2}\theta } \right) \times \left( {\sin \theta - {\text{cosec}}\,\theta } \right)\left( {\cos \theta - \sec \theta } \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \left( {\frac{{{{\sin }^2}\theta - 1}}{{\sin \theta }}} \right)\left( {\frac{{{{\cos }^2}\theta - 1}}{{\cos \theta }}} \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \left( {\frac{{ - {{\cos }^2}\theta }}{{\sin \theta }}} \right)\left( {\frac{{ - {{\sin }^2}\theta }}{{\cos \theta }}} \right) \cr & = \left( {{\text{cose}}{{\text{c}}^2}\theta } \right)\left( {{{\sec }^2}\theta } \right) \times \sin \theta \cos \theta \cr & = \sec \theta \,{\text{cosec}}\,\theta \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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