Question & Answer
QUESTION

Prove that the given equation ${{\tan }^{2}}A-{{\sin }^{2}}A={{\tan }^{2}}A.si{{n}^{2}}A$.

ANSWER Verified Verified
Hint: We will first find the relation between $\tan A,\sin A$ and $\cos A$. And then substitute the value of $\tan A$ with that relation. Further we will find a relation between $\sin A$ and $\cos A$, and use that in the equation to obtain the required expression.

Complete Step-by-Step solution:
Consider a right-angled triangle ABC, right angled at B. Let $A$ be the angle formed in the triangle at vertex A. Also, let perpendicular side with respect to angle $A$ be $p$, base with respect to angle $A$ be $b$ and hypotenuse be $h$.
Then, from the definition of $\sin A,\cos A \text{and}\tan A$, we have,
$\sin A=\dfrac{p}{h}$,
$\cos A=\dfrac{b}{h}$,
and, $\tan A=\dfrac{p}{b}$.
Now, we try to form a relation between $\tan A,\sin A$ and $\cos A$. For this, let us divide $\sin \theta $ and $\cos \theta $. So, we get,
\[\dfrac{\sin A}{\cos A}=\dfrac{\dfrac{p}{h}}{\dfrac{b}{h}}\]
\[=\dfrac{p}{h}\div \dfrac{b}{h}\]
Converting division to multiplication by taking reciprocal of the term in above equation, we get,
\[=\dfrac{p}{h}\times \dfrac{h}{b}\]
Cancelling $h$ from numerator and denominator, we get,
\[=\dfrac{p}{b}\]
\[=\tan A\]
Therefore, $\tan A=\dfrac{\sin A}{\cos A}\cdots \cdots \left( i \right)$
Again, to form a relation between $\sin A$ and $\cos A$, let us square both the terms and add them. So, we get,
${{\sin }^{2}}A+{{\cos }^{2}}A={{\left( \dfrac{p}{h} \right)}^{2}}+{{\left( \dfrac{b}{h} \right)}^{2}}$
 $=\dfrac{{{p}^{2}}}{{{h}^{2}}}+\dfrac{{{q}^{2}}}{{{h}^{2}}}$
Taking LCM,
$=\dfrac{{{p}^{2}}+{{q}^{2}}}{{{h}^{2}}}$
Using Pythagoras theorem, which states that: Square of the hypotenuse of a right angled triangle is equal to sum of square of remaining two sides of a triangle , that is, ${{p}^{2}}+{{b}^{2}}={{h}^{2}}$; in above equation, we get,
$\dfrac{{{p}^{2}}+{{b}^{2}}}{{{h}^{2}}}=\dfrac{{{h}^{2}}}{{{h}^{2}}}=1$
$\begin{align}
  & \Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
 & \Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\cdots \cdots \left( ii \right) \\
\end{align}$
Now, taking left hand side of the given equation, we have,
${{\tan }^{2}}A-{{\sin }^{2}}A$
Using equation $\left( i \right)$ here, we get,
$\begin{align}
  & {{\tan }^{2}}A-{{\sin }^{2}}A \\
 & ={{\left( \dfrac{\sin A}{\cos A} \right)}^{2}}-{{\sin }^{2}}A \\
 & =\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A \\
\end{align}$
Taking L.C.M. of ${{\cos }^{2}}A$ and 1, which is ${{\cos }^{2}}A$, we get,
$\begin{align}
  & =\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-\dfrac{{{\cos }^{2}}A\times {{\sin }^{2}}A}{{{\cos }^{2}}A} \\
 & =\dfrac{{{\sin }^{2}}A-{{\cos }^{2}}A\times {{\sin }^{2}}A}{{{\cos }^{2}}A} \\
\end{align}$
Taking ${{\sin }^{2}}A$ common from the numerator, we get,
$=\dfrac{{{\sin }^{2}}A(1-{{\cos }^{2}}A)}{{{\cos }^{2}}A}$
Using equation $\left( ii \right)$ here, we get,
$\begin{align}
  & \dfrac{{{\sin }^{2}}A(1-{{\cos }^{2}}A)}{{{\cos }^{2}}A} \\
 & =\dfrac{{{\sin }^{2}}A.{{\sin }^{2}}A}{{{\cos }^{2}}A} \\
 & ={{\left( \dfrac{\sin A}{\cos A} \right)}^{2}}{{\sin }^{2}}A \\
\end{align}$
Using equation $\left( i \right)$ here, we get,
$\begin{align}
  & ={{\left( \dfrac{\sin A}{\cos A} \right)}^{2}}{{\sin }^{2}}A \\
 & ={{\tan }^{2}}A.{{\sin }^{2}}A \\
\end{align}$
Therefore, ${{\tan }^{2}}A-{{\sin }^{2}}A={{\tan }^{2}}A.si{{n}^{2}}A$.
Hence, the given equation is proved.

Note: In this question, you can also start by applying ${{\tan }^{2}}A+{{\sec }^{2}}A=1$ identity. But it will require extra calculations. So, it is better to take this approach.