Question

Prove the following:
\[\dfrac{1-\sin A\cos A}{\cos A\left( \sec A-\operatorname{cosec}A \right)}.\dfrac{{{\sin }^{2}}A-{{\cos }^{2}}A}{{{\sin }^{3}}A+{{\cos }^{3}}A}=\sin A\]

Answer 1

Hint: To prove this expression, we should know a few trigonometric identities like \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\], \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]. Also, we should know a few algebraic formulas like \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\], \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. By using these formulas, we can prove the desired result.

Complete step-by-step solution -
In this question, we have to prove that
\[\dfrac{1-\sin A\cos A}{\cos A\left( \sec A-\operatorname{cosec}A \right)}.\dfrac{{{\sin }^{2}}A-{{\cos }^{2}}A}{{{\sin }^{3}}A+{{\cos }^{3}}A}=\sin A\]
To prove this expression, we will first consider the left hand side of the expression. So, we can write it as
\[LHS=\dfrac{1-\sin A\cos A}{\cos A\left( \sec A-\operatorname{cosec}A \right)}.\dfrac{{{\sin }^{2}}A-{{\cos }^{2}}A}{{{\sin }^{3}}A+{{\cos }^{3}}A}\]
Now, we know that \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]. So, we can write the LHS as,
\[LHS=\dfrac{1-\sin A\cos A}{\cos A\left( \sec A-\operatorname{cosec}A \right)}.\dfrac{\left( \sin A-\cos A \right)\left( \sin A+\cos A \right)}{{{\sin }^{3}}A+{{\cos }^{3}}A}\]
Now, we also know that \[{{a}^{3}}+{{b}^{3}}\] can be represented as \[\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. So, we can write LHS as,
\[LHS=\dfrac{1-\sin A\cos A}{\cos A\left( \sec A-\operatorname{cosec}A \right)}.\dfrac{\left( \sin A-\cos A \right)\left( \sin A+\cos A \right)}{\left( \sin A+\cos A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A \right)}\]
Now, we can see that (sin A + cos A) is common in both numerator and denominator of LHS. So, after canceling them out, we will get the LHS as,
\[LHS=\dfrac{\left( 1-\sin A\cos A \right)\left( \sin A-\cos A \right)}{\cos A\left( \sec A-\operatorname{cosec}A \right)\left( {{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A \right)}\]
Now, we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So, we can write
\[{{\sin }^{2}}A+{{\cos }^{2}}A-\sin A\cos A=1-\sin A\cos A\]
Therefore, we can write the LHS as,
\[LHS=\dfrac{\left( 1-\sin A\cos A \right)\left( \sin A-\cos A \right)}{\cos A\left( \sec A-\operatorname{cosec}A \right)\left( 1-\sin A\cos A \right)}\]
Now, we can see that (1 – sin A cos A) is common in both the numerator and denominator. So, we can cancel them. Therefore, we will get,
\[LHS=\dfrac{\left( \sin A-\cos A \right)}{\cos A\left( \sec A-\operatorname{cosec}A \right)}\]
Now, we know that \[\cos A=\dfrac{1}{\sec A}\] and \[\sin A=\dfrac{1}{\operatorname{cosec}A}\]. So we can write it as \[\sec A=\dfrac{1}{\cos A}\] and \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]. Therefore, we will get LHS as,
\[LHS=\dfrac{\sin A-\cos A}{\cos A\left( \dfrac{1}{\cos A}-\dfrac{1}{\sin A} \right)}\]
Now, we will take the LCM of the term in the denominator. So, we will get,
\[LHS=\dfrac{\sin A-\cos A}{\cos A\left( \dfrac{\sin A-\cos A}{\sin A\cos A} \right)}\]
\[LHS=\dfrac{\left( \sin A-\cos A \right)\left( \sin A\cos A \right)}{\left( \cos A \right)\left( \sin A-\cos A \right)}\]
Now, we can see that (cos A) and (sin A – cos A) are common in both the numerator and denominator. So, we cancel them. Therefore, we will get LHS as,
LHS = sin A
LHS = RHS
Hence proved.

Note: The possible mistakes one can make while proving the desired result is either a calculation mistake or by putting the wrong sign in the formula. And there are also possibilities that one might get confused with the question. So, solve the question carefully.