Question

Prove that:
${{\tan }^{2}}A-{{\sin }^{2}}A={{\sin }^{4}}A{{\sec }^{2}}A$

Answer 1

We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Convert the given expressions in terms of $\sin A$ and $\cos A$ by using the definition $\tan \theta =\dfrac{sin\theta }{\cos \theta }$.
We can convert $\sec A$ in terms of $\cos A$ by using the definition $\sec \theta =\dfrac{1}{\cos \theta }$.

Complete step by step answer:
Let us look at the LHS and RHS of the given identity one by one.
LHS = ${{\tan }^{2}}A-{{\sin }^{2}}A$
Using the definition $\tan \theta =\dfrac{sin\theta }{\cos \theta }$, we get:
= $\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A$
Taking out ${{\sin }^{2}}A$ and $\dfrac{1}{{{\cos }^{2}}A}$ as common factors, we get:
= $\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\left( 1-{{\cos }^{2}}A \right)$
Using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get:
= $\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\left( {{\sin }^{2}}A+{{\cos }^{2}}A-{{\cos }^{2}}A \right)$
= $\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\left( {{\sin }^{2}}A \right)$
= $\dfrac{1}{{{\cos }^{2}}A}\times {{\sin }^{4}}A$
And,
RHS = ${{\sin }^{4}}A{{\sec }^{2}}A$
Using the definition $\sec \theta =\dfrac{1}{\cos \theta }$, we get:
= ${{\sin }^{4}}A\times \dfrac{1}{{{\cos }^{2}}A}$

Since LHS = RHS, hence proved.

Note: In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
$\sin \theta =\dfrac{P}{H},\cos \theta =\dfrac{B}{H},\tan \theta =\dfrac{P}{B}$
${{P}^{2}}+{{B}^{2}}={{H}^{2}}$ (Pythagoras' Theorem)
The identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$ and $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta$ are equivalent to each other and they are a direct result of the Pythagoras' theorem.