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
\[\begin{array}{*{20}{l}}
{\therefore \sec \theta + {{\tan }^3}\theta \cos ec\theta }\\
{ = \dfrac{1}{{\cos \theta }} + {{\tan }^2}\theta \times \dfrac{1}{{\sin \theta }} \times \dfrac{{\sin \theta }}{{\cos \theta }}}\\
{ = \dfrac{1}{{\cos \theta }} + {{\tan }^2}\theta \times \dfrac{1}{{\cos \theta }}}\\
{ = \dfrac{{1 + {{\tan }^2}\theta }}{{\cos \theta }}}\\
{}
\end{array}\]
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.