Question

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

Answer 1

Hint: 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.