Question

How to establish this identity $\dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \cos x - \sin x$ ?

Answer 1

Identity is used to generalize a relation, so that it can be used in other big problems to solve them easily, so to establish this identity we have to show that the part on the right side of the equal to sign is equal to the part on the left side of the equal to sign. For that, we will take any one side and solve it using the trigonometric ratios or identities and make it equal to the other side.

Complete step-by-step solution:
Let us solve the left-hand side,
We know that $\cos ( - x) = \cos x$ and $\sin ( - x) = - \sin x$
So, we get –

$$\Rightarrow \dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \dfrac{{{{( - \sin x)}^2} - {{(\cos x)}^2}}}{{ - \sin x - \cos x}} \\\ \Rightarrow \dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{ - (\sin x + \cos x)}} \\\$$

We know that ${a^2} - {b^2} = (a - b)(a + b)$ , so –

$$\Rightarrow \dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \dfrac{{(\sin x - \cos x)(\sin x + \cos x)}}{{ - (\sin x + \cos x)}} \\\ \Rightarrow \dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = - (\sin x - \cos x) \\\ \Rightarrow \dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \cos x - \sin x \\\$$

As $\sin x + \cos x$ was present in both the numerator and the denominator, so it was canceled out. Now the left-hand side has become equal to the right-hand side.
Hence, identity is established that $\dfrac{{{{\sin }^2}( - x) - {{\cos }^2}( - x)}}{{\sin ( - x) - \cos ( - x)}} = \cos x - \sin x$ .

Note: The two sides of a right-angled triangle and one angle other than the right angle are interrelated with each other by trigonometric ratios. The sine of an angle is equal to the ratio of perpendicular and the hypotenuse of the right-angled triangle and the cosine of an angle is equal to the ratio of the base and the hypotenuse of the right-angled triangle. Trigonometric ratios have different signs in different quadrants, by angle $- \theta$ we mean $360 - \theta$ so it lies in the fourth quadrant. The value of all the trigonometric ratios is negative in the fourth quadrant except the cosine function, so $\cos ( - \theta ) = \cos \theta$ while $\sin ( - \theta ) = - \sin \theta$ .