Question

Prove $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} = {\cot ^2}x $

Answer 1

Hint: To solve this question first thing we should know that $ \cos \left( {\pi + x} \right) = - \cos x $ ,
 $ \cos \left( { - x} \right) = \cos x $ , $ \sin \left( {\pi - x} \right) = \sin x $ , $ \cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x $ , after this we proceed accordingly.

Complete step-by-step answer:
Given, $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} = {\cot ^2}x $ .
Now, consider the L.H.S of the equation.
 $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} $
We can observe there is a lot of trigonometric identity which leads to other results, as,
 $ \Rightarrow \cos \left( {\pi + x} \right) = - \cos x $
 $ \Rightarrow \cos \left( { - x} \right) = \cos x $
 $ \Rightarrow \sin \left( {\pi - x} \right) = \sin x $
 $ \Rightarrow \cos \left( {\dfrac{\pi }{2} + x} \right) = - \sin x $
Now, if we substitute $ - \cos x $ for $ \cos \left( {\pi + x} \right) $ , $ \cos x $ for $ \cos \left( { - x} \right) $ , $ \sin x $ for $ \sin \left( {\pi - x} \right) $ and $ - \sin x $ for $ \cos \left( {\dfrac{\pi }{2} + x} \right) $ in the expression $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} $ .
 $
\Rightarrow \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} = \dfrac{{ - \cos x\left( {\cos x} \right)}}{{\sin x\left( { - \sin x} \right)}} \\
   = \dfrac{{ - {{\cos }^2}x}}{{ - {{\sin }^2}x}} \\
   = {\cot ^2}x \;
  $
So, we can conclude that on simplification of the expression
 $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} $ we get the result equal to $ {\cot ^2}x $ .
Hence, proved
 $ \dfrac{{\cos \left( {\pi + x} \right)\cos \left( { - x} \right)}}{{\sin \left( {\pi - x} \right)\cos \left( {\dfrac{\pi }{2} + x} \right)}} = {\cot ^2}x $ .

Note: Trigonometric identities are equalities in mathematics that include trigonometric functions that are valid for each value of the variables occurring where both sides of the equality are defined. There are identities, geometrically, containing the functions of one or more angles.The main description of trigonometry functions is the angles of sine, cosine, and tangent. And from the main functions can be derived the three functions that are cotangent, secant, and cosecant. Basically, in contrast to the main trigonometric functions, the other three functions are also used.
The cos and sec functions are even functions; the rest other functions are odd functions.