Question

How do you simplify $\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ ?

Answer 1

Hint : We first simplify the numerator and the denominator part of $\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ separately. We use the identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . We also know that the terms $\sin x$ and $\csc x$ are inverse of each other which gives $\sin x\times \csc x=1$ . For both parts we get the value of the simplification but the signs are opposite to each other. The final answer becomes $-1$ .

Complete step-by-step answer :
We have been given a trigonometrical fraction of $\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ .
We apply different theorems of identity for the numerator and the denominator of $\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ .
We simplify the numerator part to get
$1-{{\left( \sin x \right)}^{2}}=1-{{\sin }^{2}}x$ .
For the numerator part we have
$1-{{\sin }^{2}}x={{\cos }^{2}}x$ as for any value of $x$ , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ .
For the denominator part we multiply $\sin x$ to both terms of $\sin x-\csc x$ .
We know that the terms $\sin x$ and $\csc x$ are inverse of each other.
So,
$\sin x\times \csc x=1$ as $\sin x=\dfrac{1}{\csc x}$ .
After the multiplication we get
$\sin x\left( \sin x-\csc x \right)={{\sin }^{2}}x-\sin x\csc x$ .
We put the values to get
$\sin x\left( \sin x-\csc x \right)={{\sin }^{2}}x-1$ .
We again put the identity value of
$1-{{\sin }^{2}}x={{\cos }^{2}}x$ to get $\sin x\left( \sin x-\csc x \right)=-{{\cos }^{2}}x$ .
Now we put all the values to get the expression
$\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ as
$\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}=\dfrac{{{\cos }^{2}}x}{-{{\cos }^{2}}x}=-1\times \sin x$
Therefore, the simplified solution of
$\dfrac{1-{{\left( \sin x \right)}^{2}}}{\sin x-\csc x}$ is $-\sin x$ .
So, the correct answer is “ $-\sin x$”.

Note : We need to remember that the final terms are square terms. The identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\sin x\times \csc x=1$ are valid for any value of $x$ . The division of the fraction part only gives $-1$ as the solution.