Question

Simplify the expression:
$\dfrac{{\sin ({{270}^\circ} + x){{\cos }^3}({{720}^\circ} - x) - \sin ({{270}^\circ} - x){{\sin }^3}({{540}^\circ} + x)}}{{\sin ({{90}^\circ} + x)\sin ( - x) - {{\cos }^2}({{180}^\circ} - x)}} + \dfrac{{\cot ({{270}^\circ} - x)}}{{\cos e{c^2}({{450}^\circ} + x)}}$

Answer 1

Hint: In order to this question, to simplify the given expression, first we will rewrite the given expression and then we will analyse or rewrite the trigonometric ratios which we can use to simplify the given expression, and then we will simplify until we will get the simplified value.

Complete step-by-step solution:
The given expression is:
$\dfrac{{\sin ({{270}^\circ} + x){{\cos }^3}({{720}^\circ} - x) - \sin ({{270}^\circ} - x){{\sin }^3}({{540}^\circ} + x)}}{{\sin ({{90}^\circ} + x)\sin ( - x) - {{\cos }^2}({{180}^\circ} - x)}} + \dfrac{{\cot ({{270}^\circ} - x)}}{{\cos e{c^2}({{450}^\circ} + x)}}$
As we know, $\sin ({270^\circ} + \theta ) = - \cos \theta $ ,
$\sin ({270^\circ} - \theta ) = - \cos \theta $
$\cos ({720^\circ} - \theta ) = \cos \theta $ ,
$\sin ({540^\circ} + \theta ) = -\sin \theta $
$\cot ({270^\circ} - \theta ) = \tan \theta $ and
$\cos ec({450^\circ} + \theta ) = \sec \theta $
So,
$\dfrac{{\sin ({{270}^\circ} + x){{\cos }^3}({{720}^\circ} - x) - \sin ({{270}^\circ} - x){{\sin }^3}({{540}^\circ} + x)}}{{\sin ({{90}^\circ} + x)\sin ( - x) - {{\cos }^2}({{180}^\circ} - x)}} + \dfrac{{\cot ({{270}^\circ} - x)}}{{\cos e{c^2}({{450}^\circ} + x)}}$
$= \dfrac{{ - \cos x \times {{\cos }^3}x - (-\cos x \times {-{\sin }^3}x)}}{{ - \cos x\sin x - {{\cos }^2}x}} + \dfrac{{\tan x}}{{{{\sec }^2}x}} $
$= \dfrac{{ - {{\cos }^4}x - {{\sin }^3}x}\cos x}{{ - \cos x\sin x - {{\cos }^2}x}} + \dfrac{{\sin x}}{{\cos x}} \times {\cos ^2}x $
$= \dfrac{{ -\cos x ({{\cos }^3}x + {{\sin }^3}x)}}{{ - \cos x(\sin x + \cos x)}} + \sin x\cos x $
$=\dfrac{{{{{\sin}^3}x}+{{\cos }^3}x}}{{\sin x+\cos x}} + \sin x\cos x $
We know that ${a}^3+{b}^3 = \left(a+b\right)\left(a^2-ab+b^2\right)$ on using this identity
$=\dfrac{\require{\cancel}{\cancel{\left(\sin x+\cos x \right)}}\left({{\cos}^2{x}}-\cos x \sin x +{{\sin}^{2}x}\right)}{\require{\cancel}{\cancel{{\sin x+\cos x}}}}+\sin x \cos x$
$={{\cos }^2}x - \sin x\cos x +{{\sin }^2}x+\sin x \cos x $
$= {{\cos }^2}x+{{\sin }^2}x$
 $=1$
Since we know the identity ${{\cos }^2}x+{{\sin }^2}x=1 $. Hence the final answer is 1.

Note:
$\bullet $ There are multiple ways to represent a trigonometric expression.
$\bullet$ Simplifying one side of the equation to equal the other side is a method for verifying an identity.
$ \bullet $ The approach to verifying an identity depends on the nature of the identity.
$ \bullet $ We can create an identity from a given expression.
$ \bullet $ Verifying an identity may involve algebra with the fundamental identities.