Question

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

Answer 1

Hint:
Considering the left hand side , first taking $\sec \theta $ common and replacing$\sec \theta = \dfrac{1}{{\cos \theta }}{\text{ and }}\cos ec\theta = \dfrac{1}{{\sin \theta }}$ we get an equation in which we can further replace ${\tan ^3}\theta = \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}$ and then by using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$ we can find the value of $\sec \theta $. And using the given statement ${\tan ^2}\theta = 1 - {a^2}$ we get our right hand side.

Complete step by step solution:
Here we are give that ${\tan ^2}\theta = 1 - {a^2}$……….(1)
We are asked to prove $\sec \theta + {\tan ^3}\theta \cos ec\theta = {\left( {2 - {a^2}} \right)^{\dfrac{3}{2}}}$
So now lets start with our left hand side
$ \Rightarrow \sec \theta + {\tan ^3}\theta \cos ec\theta $
At first lets multiply and divide by $\sec \theta $
 $
   \Rightarrow \dfrac{{\sec \theta }}{{\sec \theta }}\left[ {\sec \theta + {{\tan }^3}\theta \cos ec\theta } \right] \\
   \Rightarrow \sec \theta \left[ {\dfrac{{\sec \theta + {{\tan }^3}\theta \cos ec\theta }}{{\sec \theta }}} \right] \\
   \Rightarrow \sec \theta \left[ {1 + \dfrac{{{{\tan }^3}\theta \cos ec\theta }}{{\sec \theta }}} \right] \\
 $
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}{\text{ and }}\cos ec\theta = \dfrac{1}{{\sin \theta }}$
Using this in the above equation we get
$
   \Rightarrow \sec \theta \left[ {1 + \dfrac{{{{\tan }^3}\theta \dfrac{1}{{\sin \theta }}}}{{\dfrac{1}{{\cos \theta }}}}} \right] \\
   \Rightarrow \sec \theta \left[ {1 + {{\tan }^3}\theta \dfrac{{\cos \theta }}{{\sin \theta }}} \right] \\
 $
Once again since $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ we can write ${\tan ^3}\theta = \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}$
Using this in the above equation we get
$
   \Rightarrow \sec \theta \left[ {1 + \dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}\dfrac{{\cos \theta }}{{\sin \theta }}} \right] \\
   \Rightarrow \sec \theta \left[ {1 + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right] \\
   \Rightarrow \sec \theta \left[ {1 + {{\tan }^2}\theta } \right] \\
 $
Let the above equation be (2)
we know the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$
from this we can get the value of $\sec \theta $
$
   \Rightarrow {\sec ^2}\theta = 1 + {\tan ^2}\theta \\
   \Rightarrow \sec \theta = \sqrt {1 + {{\tan }^2}\theta } \\
 $
Now let's use this in equation (2)
$
   \Rightarrow \sqrt {1 + {{\tan }^2}\theta } \left[ {1 + {{\tan }^2}\theta } \right] \\
   \Rightarrow {\left( {1 + {{\tan }^2}\theta } \right)^{\dfrac{1}{2}}}\left[ {1 + {{\tan }^2}\theta } \right] \\
 $
Since the base is the same we can add up the powers
$ \Rightarrow {\left( {1 + {{\tan }^2}\theta } \right)^{\dfrac{3}{2}}}$
From equation (1) we have ${\tan ^2}\theta = 1 - {a^2}$
Using this we get
$
   \Rightarrow {\left( {1 + \left( {1 - {a^2}} \right)} \right)^{\dfrac{3}{2}}} \\
   \Rightarrow {\left( {1 + 1 - {a^2}} \right)^{\dfrac{3}{2}}} \\
   \Rightarrow {\left( {2 - {a^2}} \right)^{\dfrac{3}{2}}} \\
 $
Hence we have obtained the right hand side .
Hence proved

Note:
Steps to keep in mind while solving trigonometric problems are
1) Always start from the more complex side
2) Express everything into sine and cosine
3) Combine terms into a single fraction
4) Use Pythagorean identities to transform between ${\sin ^2}x{\text{ and }}{\cos ^2}x$
5) Know when to apply double angle formula
6) Know when to apply addition formula
7) Good old expand/ factorize/ simplify/ cancelling