Question

$$\begin{gathered} \text{If sin}\theta \text{, cos}\theta \text{ and tan}\theta \text{ are in G}\text{.P}\text{. then co}{\text{t}}^6\theta -{\cot }^2\theta \text{ is} \\ \text{A}\text{. 1} \\ \text{B}\text{. }{\displaystyle \frac{1}{2}} \\ \text{C}\text{. 2} \\ \text{D}\text{. 3} \end{gathered}$$

Answer 1

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