Question

If the value of ${\tan ^2}\theta = 1 - {e^2},$ then prove that $\sec \theta + {\tan ^3}\theta \cos ec\theta = {\left( {2 - {e^2}} \right)^{\dfrac{3}{2}}}$

Answer 1

Hint: First solve it in trigonometric function and then apply the value of ${\tan ^2}\theta$ you will also need to find the value of $\sec \theta$ and avoid getting the value of ${\tan ^3}\theta \& \cos ec\theta$
Complete Step by Step Solution:
We are given that ${\tan ^2}\theta = 1 - {e^2}$
Let us try to solve the function we are told to prove

{\therefore \sec \theta + {{\tan }^3}\theta \cos ec\theta }\\\ { = \dfrac{1}{{\cos \theta }} + {{\tan }^2}\theta \times \dfrac{1}{{\sin \theta }} \times \dfrac{{\sin \theta }}{{\cos \theta }}}\\\ { = \dfrac{1}{{\cos \theta }} + {{\tan }^2}\theta \times \dfrac{1}{{\cos \theta }}}\\\ { = \dfrac{{1 + {{\tan }^2}\theta }}{{\cos \theta }}}\\\ {} \end{array}$$ So let us try to find the value of $$\sec \theta $$ We know that $$\begin{array}{*{20}{l}} {\therefore {{\sec }^2}\theta = 1 + {{\tan }^2}\theta }\\\ { \Rightarrow {{\sec }^2}\theta = 1 + 1 - {e^2}}\\\ { \Rightarrow {{\sec }^2}\theta = 2 - {e^2}}\\\ { \Rightarrow \sec \theta = \sqrt {2 - {e^2}} }\\\ {} \end{array}$$ Substituting $${{{\sec }^2}\theta = 1 + {{\tan }^2}\theta }$$ and we will get $${\dfrac{{{{\sec }^2}\theta }}{{\cos \theta }} = {{\sec }^3}\theta }$$ We will get it as $$\begin{array}{l} \therefore {\left( {\sec \theta } \right)^3}\\\ = {\left( {\sqrt {2 - {e^2}} } \right)^3}\\\ = {\left[ {{{\left( {2 - {e^2}} \right)}^{\dfrac{1}{2}}}} \right]^3}\\\ = {\left( {2 - {e^2}} \right)^{\dfrac{3}{2}}} \end{array}$$ Hence Proved Note: We can also do it by finding the values of $$\sin \theta \& \cos \theta $$ and then $$\sec \theta \& \cos ec\theta $$ but that will be a long process and the chances of making mistakes are higher better to solve the trigonometric functions as much as you can before putting the final value.