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

Last updated date: 15th Jul 2024
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Hint: Expand the L.H.S term and simplify it and Use Trigonometric ratios and identities to get the required 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}
\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}