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Question: How do you simplify the expression \(\dfrac{1-\csc x}{1-{{\csc }^{2}}x}?\)...

How do you simplify the expression 1−csc⁡x1−csc⁡2x?\dfrac{1-\csc x}{1-{{\csc }^{2}}x}?

Explanation

Solution

The above given question is of trigonometric identities. So, we will first convert all the csc⁡x\csc x into sin⁡x\sin x terms by using the formula csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x} and then we will use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 and write 1−cos⁡2x=sin⁡2x1-{{\cos }^{2}}x={{\sin }^{2}}x and then we will simplify the given expression.

Complete answer:
We can see that above given question is of trigonometric identity and so we will use trigonometric identity to simplify 1−csc⁡x1−csc⁡2x\dfrac{1-\csc x}{1-{{\csc }^{2}}x}.
At first, we will convert all the csc⁡x\csc x into sin⁡x\sin x terms by using the formula csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}, then we will get:
⇒1−csc⁡x1−csc⁡2x=1−1sin⁡x1−1sin⁡2x\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{1-\dfrac{1}{\sin x}}{1-\dfrac{1}{{{\sin }^{2}}x}}
⇒1−csc⁡x1−csc⁡2x=1−1sin⁡x1−1sin⁡2x\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{1-\dfrac{1}{\sin x}}{1-\dfrac{1}{{{\sin }^{2}}x}}
⇒1−csc⁡x1−csc⁡2x=sin⁡x−1sin⁡xsin⁡2x−1sin⁡2x\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{\dfrac{\sin x-1}{\sin x}}{\dfrac{{{\sin }^{2}}x-1}{{{\sin }^{2}}x}}
⇒1−csc⁡x1−csc⁡2x=(sin⁡x−1)sin⁡2x(sin⁡2x−1)sin⁡x\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{\left( \sin x-1 \right){{\sin }^{2}}x}{\left( {{\sin }^{2}}x-1 \right)\sin x}
⇒1−csc⁡x1−csc⁡2x=(sin⁡x−1)sin⁡x(sin⁡2x−1)\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{\left( \sin x-1 \right)\sin x}{\left( {{\sin }^{2}}x-1 \right)}
Now, we will use the formula (a2−b2)=(a−b)(a+b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) so, we can write sin⁡2x−1=(sin⁡x−1)(sin⁡x+1){{\sin }^{2}}x-1=\left( \sin x-1 \right)\left( \sin x+1 \right).
⇒1−csc⁡x1−csc⁡2x=(sin⁡x−1)sin⁡x(sin⁡x−1)(sin⁡x+1)\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{\left( \sin x-1 \right)\sin x}{\left( \sin x-1 \right)\left( \sin x+1 \right)}
Now, after cancelling the common terms from numerator and denominator, we will get:
⇒1−csc⁡x1−csc⁡2x=sin⁡x(sin⁡x+1)\Rightarrow \dfrac{1-\csc x}{1-{{\csc }^{2}}x}=\dfrac{\sin x}{\left( \sin x+1 \right)}
So, the simplified form of 1−csc⁡x1−csc⁡2x\dfrac{1-\csc x}{1-{{\csc }^{2}}x} is sin⁡x1+sin⁡x\dfrac{\sin x}{1+\sin x}.
This is our required solution.

Note: Students are required to note that they should memorize all the trigonometric formulas otherwise they will not be able to solve trigonometric formulas. Also, note that when we have simplified the given trigonometric expression we usually convert all the terms in the given expression into sin x and cos x and then simplify it.