Question

Prove the following:
\[\dfrac{{{\sin }^{3}}A+{{\cos }^{3}}A}{\sin A+\cos A}+\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A-\cos A}=2\]

Answer 1

Hint: In order to prove this expression, we should have some knowledge of the trigonometric identities like \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] and few algebraic identities like \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] and \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. By using these formulas, we will be able to prove the required results.

Complete step-by-step solution -
In the given question, we have to prove that
\[\dfrac{{{\sin }^{3}}A+{{\cos }^{3}}A}{\sin A+\cos A}+\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A-\cos A}=2\]
To solve this question, we will first consider the left hand side of the expression. So, we can write it as,
\[LHS=\dfrac{{{\sin }^{3}}A+{{\cos }^{3}}A}{\sin A+\cos A}+\dfrac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A-\cos A}\]
Now, we know that \[{{a}^{3}}+{{b}^{3}}\] can be expressed as \[\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] and \[{{a}^{3}}-{{b}^{3}}\] can be expressed as \[\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. So, we can write\[{{\sin }^{3}}A+{{\cos }^{3}}A=\left( \sin A+\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A \right)\]
And \[{{\sin }^{3}}A-{{\cos }^{3}}A=\left( \sin A-\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\sin A\cos A \right)\]
Therefore, we will get LHS as
\[LHS=\dfrac{\left( \sin A+\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A \right)}{\sin A+\cos A}+\dfrac{\left( \sin A-\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\sin A\cos A \right)}{\sin A-\cos A}\]
Now, we know that the common terms in the numerator and denominator get canceled out. Therefore, we will get LHS as,
\[LHS=\left( {{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A \right)+\left( {{\sin }^{2}}A+{{\cos }^{2}}A+\sin A\cos A \right)\]
Now, we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] and we also know that the same terms with opposite signs get canceled out. So, we will get the LHS as
LHS = 1 + 1
LHS = 2
Therefore, LHS = RHS
Hence proved.

Note: In this question, the possible mistake one can do is by taking the LCM and then simplifying it. Because when we take the LCM, the number of terms in the numerator will increase and we might end up with some wrong calculation and complicated solution.