Question

Prove the following statement.
$\dfrac{\cos A\operatorname{cosec}A-\sin A\sec A}{\cos A+\sin A}=\operatorname{cosec}A-\sin A$

Answer 1

Hint: In order to prove the given relation, we should have some knowledge regarding the concepts of trigonometry, like the trigonometric formulas, $\operatorname{cosec}A=\dfrac{1}{\sin A}$ and $\sec A=\dfrac{1}{\cos A}$. Also, we should know the algebraic property that, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. By using these properties, we will get the desired result.
Complete step-by-step answer:
In this question, we have been asked to prove $\dfrac{\cos A\operatorname{cosec}A-\sin A\sec A}{\cos A+\sin A}=\operatorname{cosec}A-\sin A$. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\dfrac{\cos A\operatorname{cosec}A-\sin A\sec A}{\cos A+\sin A}$
Now, we know that cosec A can be written as $\dfrac{1}{\sin A}$ and sec A can be written as $\dfrac{1}{\cos A}$. So, we get,
$\begin{align}
  & LHS=\dfrac{\cos A\dfrac{1}{\sin A}-\sin A\dfrac{1}{\cos A}}{\cos A+\sin A} \\
 & LHS=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\cos A}}{\cos A+\sin A} \\
\end{align}$
Now, we will take the LCM of the terms in the numerator. So, we will get,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\sin A\cos A}}{\cos A+\sin A}$
And we can further write it as,
$LHS=\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\left( \sin A\cos A \right)\left( \cos A+\sin A \right)}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. So, we can write ${{\cos }^{2}}A-{{\sin }^{2}}A=\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)$. Therefore, we can write the LHS as follows.
$LHS=\dfrac{\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)}{\left( \sin A\cos A \right)\left( \cos A+\sin A \right)}$
Here, we can see that (cos A + sin A) is common in both the numerator and the denominator. So, we can cancel them. Hence, the LHS will be,
$LHS=\dfrac{\left( \cos A-\sin A \right)}{\sin A\cos A}$
We can further write it as,
$LHS=\dfrac{\cos A}{\sin A\cos A}-\dfrac{\sin A}{\sin A\cos A}$
We know that the common terms in the numerator and the denominator can be cancelled out. So, we will get,
$LHS=\dfrac{1}{\sin A}-\dfrac{1}{\cos A}$
We know that $\dfrac{1}{\sin A}=\operatorname{cosec}A$ and $\dfrac{1}{\cos A}=\sec A$. Therefore, we can write the LHS as,
$LHS=\operatorname{cosecA}-\sec A$
$LHS=RHS$
Hence proved.

Note: We can also solve this question by starting from the RHS, by multiplying and dividing each term with (cos A + sin A) and then later at the time of simplifying, we will put sin A cosec A = 1 because $\sin A=\dfrac{1}{\operatorname{cosec}A}$ and cos A sec A = 1 because $\cos A=\dfrac{1}{\sec A}$ and then we will get our answer.