Question

Prove that: ${{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta \times {{\sin }^{2}}\theta $

Answer 1

Hint: Simplify the terms given in the LHS of the equation by writing $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and then by using$\sec \theta =\dfrac{1}{\cos \theta }$. After that, use ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to simplify the LHS and get LHS equal to RHS.

Complete step-by-step answer:
We have to just simplify the LHS to get the expression given in RHS.

Taking, $LHS={{\tan }^{2}}\theta -{{\sin }^{2}}\theta $

Using \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] in the above expression we get-

$LHS={{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}-{{\sin }^{2}}\theta $

$LHS=\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta $

$LHS={{\sin }^{2}}\theta \left[ \dfrac{1}{{{\cos }^{2}}\theta }-1 \right]$

Taking \[{{\sin }^{2}}\theta \] (common from both terms).

Since, we know that $\sec \theta =\dfrac{1}{\cos \theta }$

That means, ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }\left( on\,squaring\,both\,sides \right)$.

Using, $\dfrac{1}{{{\cos }^{2}}\theta }={{\sec }^{2}}\theta $, in the above expression we get-

$LHS={{\sin }^{2}}\theta \left\{ {{\sec }^{2}}\theta -1 \right\}$

Now, as we know that, $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $

i.e. ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$

using, $\left( {{\sec }^{2}}\theta -1 \right)={{\tan }^{2}}\theta $ in the above expression we get-

$\begin{align}

  & LHS=\left( {{\sin }^{2}}\theta \right)\left( {{\tan }^{2}}\theta \right) \\

 & LHS={{\tan }^{2}}\theta \times {{\sin }^{2}}\theta \\

 & LHS=RHS \\

\end{align}$

Hence, ${{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta \times {{\sin }^{2}}\theta (proved)$


Note: Students can also solve this by this method given as-

We are given, $LHS={{\tan }^{2}}\theta -{{\sin }^{2}}\theta $

\[\begin{align}

  & LHS={{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}-{{\sin }^{2}}\theta \\

 & LHS=\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta \\

 & LHS=\dfrac{{{\sin }^{2}}\theta -\left( {{\sin }^{2}}\theta \right)\left( {{\cos }^{2}}\theta \right)}{{{\cos }^{2}}\theta } \\

 & LHS=\dfrac{{{\sin }^{2}}\theta \times \left( 1-{{\cos }^{2}}\theta \right)}{{{\cos }^{2}}\theta } \\

 & LHS={{\sin }^{2}}\theta \left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }

\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\,i\,.e.\,1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \right] \\

 & LHS={{\sin }^{2}}\theta \left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)\, \\

 & LHS={{\sin }^{2}}\theta \times {{\left( \tan \theta \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \dfrac{\sin \theta }{\cos \theta }=\tan \theta \right] \\

 & LHS={{\sin }^{2}}\theta \times {{\tan }^{2}}\theta \\

 & \\

\end{align}\]

Hence, LHS=RHS

Therefore, ${{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta \times {{\sin }^{2}}\theta \,\,(proved)$