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.