$$\left( {\frac{{{{\tan }^3}\theta }}{{{{\sec }^2}\theta }} + \frac{{{{\cot }^3}\theta }}{{{\text{cose}}{{\text{c}}^2}\theta }} + 2\sin \theta \cos \theta } \right)$$ ÷ (1 + cosec2θ + tan2θ), 0° < θ < 90°, is equal to:
A. cosecθ
B. cosecθsecθ
C. sinθcosθ
D. secθ
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \left( {\frac{{{{\tan }^3}\theta }}{{{{\sec }^2}\theta }} + \frac{{{{\cot }^3}\theta }}{{{\text{cose}}{{\text{c}}^2}\,\theta }} + 2\sin \theta \cos \theta } \right) \div \left( {1 + {\text{cose}}{{\text{c}}^2}\theta + {{\tan }^2}\theta } \right) \cr & = \left( {\frac{{{{\sin }^3}\theta }}{{\cos \theta }} + \frac{{{{\cos }^3}\theta }}{{{\text{sin}}\,\theta }} + 2\sin \theta \cos \theta } \right) \div \left( {{\text{cose}}{{\text{c}}^2}\theta + {{\sec }^2}\theta } \right) \cr & = \left( {\frac{{{{\sin }^4}\theta + {{\cos }^4}\theta + 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}} \right) \div \left( {\frac{1}{{{{\sin }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right) \cr & = \left( {\frac{{{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^2}}}{{\sin \theta \cos \theta }}} \right) \div \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right) \cr & = \left( {\frac{1}{{\sin \theta \cos \theta }}} \right) \times \left( {\frac{{{{\sin }^2}\theta {{\cos }^2}\theta }}{1}} \right) \cr & = \sin \theta \cos \theta \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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