Courses
Courses for Kids
Free study material
Free LIVE classes
More
Questions & Answers
seo-qna
LIVE
Join Vedantu’s FREE Mastercalss

Prove that ${{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right)$.

Answer
VerifiedVerified
359.7k+ views
Hint: Expand the L.H.S term and simplify it and Use Trigonometric ratios and identities to get the required answer.

“Complete step-by-step answer:”
To prove, ${{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right)$
Left hand side (L.H.S) $={{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}$
We open the brackets by the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
$\begin{align}
  & \therefore LHS=\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta }+2\tan \theta \cos \theta \right)+\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta }-2\tan \theta \cos \theta \right) \\
 & \Rightarrow LHS=2{{\tan }^{2}}\theta +\dfrac{2}{{{\cos }^{2}}\theta } \\
 & \Rightarrow LHS=2\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta } \right) \\
\end{align}$
Let us write $\tan \theta \ as\ \dfrac{\sin \theta }{\cos \theta }$.
$\begin{align}
  & \Rightarrow LHS=2\left[ {{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}+\dfrac{1}{{{\cos }^{2}}\theta } \right] \\
 & \Rightarrow LHS=2\left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }+\dfrac{1}{{{\cos }^{2}}\theta } \right) \\
 & \Rightarrow LHS=2\left( \dfrac{1+{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right) \\
\end{align}$
We know that \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \].
$\begin{align}
  & \therefore LHS=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right) \\
 & L.H.S=R.H.S \\
\end{align}$
Hence proved.

Note: In questions like these, our aim is to correct the longer side, in this case L.H.S identical to the other side. Therefore, after every step, we try to manipulate the expression.
Formulae used:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\
 & {{\tan }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \\
\end{align}$
Last updated date: 26th Sep 2023
Total views: 359.7k
Views today: 5.59k