Question

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  {\text{If sin}}\theta {\text{, cos}}\theta {\text{ and tan}}\theta {\text{ are in G}}{\text{.P}}{\text{. then co}}{{\text{t}}^6}\theta - {\cot ^2}\theta {\text{ is}} \\
  {\text{A}}{\text{. 1}} \\
  {\text{B}}{\text{. }}\dfrac{1}{2} \\
  {\text{C}}{\text{. 2}} \\
  {\text{D}}{\text{. 3}} \\
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Answer 1

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   \\
  {\text{We know that when }}a,{\text{ }}b,{\text{ }}c{\text{ are in GP then }} \\
   \Rightarrow {b^2} = a \cdot c \\
  {\text{here sin}}\theta {\text{, cos}}\theta {\text{ and tan}}\theta {\text{ are in gp}} \\
   \Rightarrow {\text{co}}{{\text{s}}^2}\theta = \sin \theta \cdot \tan \theta \\
  {\text{ = sin}}\theta \cdot \dfrac{{\sin \theta }}{{\cos \theta }} \\
   \Rightarrow \dfrac{{{\text{co}}{{\text{s}}^2}\theta }}{{{\text{si}}{{\text{n}}^2}\theta }} = \dfrac{1}{{\cos \theta }}{\text{ }} \Rightarrow \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^3}\theta }} = 1{\text{ }}........{\text{(i)}} \\
   \Rightarrow {\cot ^2}\theta = \sec \theta \\
  {\text{Now put the value of }}{\cot ^2}\theta {\text{ in the question}} \\
   \Rightarrow {\cot ^6}\theta - {\cot ^2}\theta = {\sec ^3}\theta - \sec \theta \\
  {\text{ = }}\sec \theta ({\sec ^2}\theta - 1) \\
  {\text{ = }}\sec \theta \cdot {\tan ^2}\theta \\
  {\text{ = }}\dfrac{1}{{\cos \theta }} \cdot \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} \\
  {\text{ = }}\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^3}\theta }} \\
  {\text{By putting the value in equation (i)}} \\
  {\cot ^6}\theta - {\cot ^2}\theta = 1{\text{ }} \\
  {\text{So option A is correct}}{\text{.}} \\
  {\text{Note: - Always try to use geometric mean when three consecutive term of a GP are given}}{\text{. }} \\
  {\text{these are the best method to solve the questions}}{\text{.}} \\
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