Question

How do you simplify $\left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)$ ?

Answer 1

In this question, we have been asked to simplify the given trigonometric equation. Start by converting the ratios into sin and cos. Then, find out the LCM of the denominators and make them equal. Simplify the equation that you have got now. You will see an identity being formed in the numerator. Put its value and cancel out the like terms. You will get your answer.

Formula used:

1. ${\cos ^2}x + {\sin ^2}x = 1$
2. $\dfrac{{\cos x}}{{\sin x}} = \cot x$

Complete step-by-step answer:
We are given a trigonometric expression. Let us see how to simplify it.
$\Rightarrow \left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)$ …. (given)
Simplifying the numerator by putting $\sec x = \dfrac{1}{{\cos x}}$,
$\Rightarrow \left( {\dfrac{1}{{\sin x\cos x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)$
Now, we will take LCM of the denominator. LCM will come out to be $\sin x\cos x$.
Multiplying the second term by $\sin x$ to make the denominator the same.
$\Rightarrow \left( {\dfrac{1}{{\sin x\cos x}}} \right) - \left( {\dfrac{{\sin x \times \sin x}}{{\cos x \times \sin x}}} \right)$
On simplifying, we will get,
$\Rightarrow \dfrac{{1 - {{\sin }^2}x}}{{\sin x\cos x}}$
We know that ${\cos ^2}x + {\sin ^2}x = 1$. On shifting, we will get, ${\cos ^2}x = 1 - {\sin ^2}x$. Substituting this in the above expression,
$\Rightarrow \dfrac{{{{\cos }^2}x}}{{\sin x\cos x}}$
Cancelling out the like terms,
$\Rightarrow \dfrac{{\cos x}}{{\sin x}}$
We know that $\dfrac{{\cos x}}{{\sin x}} = \cot x$.

Hence, $\left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right) = \cot x$

Note:
Let us solve the same question by another method.
$\Rightarrow \left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)$
We know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$. Putting this in the above expression,
$\Rightarrow \left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \tan x$
Now, in order to make the denominator, the same, we will multiply and divide the second term by $\sin x$.
$\Rightarrow \left( {\dfrac{{\sec x}}{{\sin x}}} \right) - \dfrac{{\tan x \times \sin x}}{{\sin x}}$
On simplifying, we will get,
$\Rightarrow \dfrac{{\sec x - \tan x \times \sin x}}{{\sin x}}$
Now, we know that $\sec x = \dfrac{1}{{\cos x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Putting them in the formula,
$\Rightarrow \dfrac{{\dfrac{1}{{\cos x}} - \dfrac{{{{\sin }^2}x}}{{\cos x}}}}{{\sin x}}$
$\Rightarrow \dfrac{{\dfrac{{1 - {{\sin }^2}x}}{{\cos x}}}}{{\sin x}}$
Now, the steps are the same from here. We know that ${\cos ^2}x + {\sin ^2}x = 1$. On shifting, we will get, ${\cos ^2}x = 1 - {\sin ^2}x$. Substituting this in the above expression,
$\Rightarrow \dfrac{{\dfrac{{{{\cos }^2}x}}{{\cos x}}}}{{\sin x}}$
Cancelling out the like terms,
$\Rightarrow \dfrac{{\dfrac{{\cos x}}{1}}}{{\sin x}}$
On simplifying, we will get, $\dfrac{{\cos x}}{{\sin x}} = \cot x$.
Hence, the answer from this method is similar to the answer from the method we used above. You can use any method from these two.