Question

Show that $\left( \text{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$.

Answer 1

Hint: In this question we have been given with a trigonometric expression for which we have to prove that the left-hand side is equal to the right-hand side. We will solve this question by first taking the left-hand side and performing trigonometric operations so that it is equal to the left-hand side. We will use the identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ to simplify the terms in the expression and get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \left( \text{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$
Consider the left-hand side of the expression, we get:
$\Rightarrow \left( \text{cosec}A-\sin A \right)\left( \sec A-\cos A \right)$
Now we know that $\text{cosec}A=\dfrac{1}{\sin A}$ and $\sec A=\dfrac{1}{\cos A}$ therefore, on substituting, we get:
$\Rightarrow \left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)$
On taking the lowest common multiple, we get:
$\Rightarrow \left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)$
Now we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ therefore, we have $1-{{\sin }^{2}}A={{\cos }^{2}}A$ and $1-{{\cos }^{2}}A={{\sin }^{2}}A$.
On substituting, we get:
$\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)$
On cancelling the terms, we get:
$\Rightarrow \dfrac{\sin A\cos A}{1}$
Now on using the expression ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, we can write the denominator as:
$\Rightarrow \dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}$
On rearranging the fractions, we get:
$\Rightarrow \dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}}$
On splitting the fraction, we get:
$\Rightarrow \dfrac{1}{\dfrac{{{\sin }^{2}}A}{\sin A\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A\cos A}}$
On cancelling the terms, we get:
$\Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}$
Now we know that $\tan A=\dfrac{\sin A}{\cos A}$ and $\cot A=\dfrac{\cos A}{\sin A}$ therefore, on substituting, we get:
$\Rightarrow \dfrac{1}{\tan A+\cot A}$, which is the right-hand side of the expression, hence proved.

Note:It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken. The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions. To simplify any given equation, it is good practice to convert all the identities into and for simplifying. If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.