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.