Question

${\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}} \\\$

Answer 1

$\\\ {\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{.}} \\\$