Question

Prove that \[\dfrac{{\cos ({{180}^{\circ}} - A)\cot ({{90}^ \circ } + A)\cos ( - A)}}{{\tan ({{180}^{\circ}} + A)\tan ({{270}^{\circ}} + A)\sin ({{360}^{\circ}} - A)}} = \cos A\]

Answer 1

Hint:
In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.

Complete step by step solution:
The below mentioned formulae may be used before solving, in solution which is as follows
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
$\cot x = \dfrac{{\cos x}}{{\sin x}}$
$\cos ecx = \dfrac{1}{{\sin x}}$
$\sec x = \dfrac{1}{{\cos x}}$
Some other important formulae that might be used in solving this question is as follows
For $\sin $ function
$
  \sin ( - x) = - \sin x \\
  \sin ({90^ \circ } - x) = \cos x \\
  \sin ({90^ \circ } + x) = \cos x \\
  \sin ({180^{\circ}} - x) = \sin x \\
  \sin ({180^{\circ}} + x) = - \sin x \\
  \sin ({270^{\circ}} - x) = - \cos x \\
  \sin ({270^{\circ}} + x) = - \cos x \\
  \sin ({360^{\circ}} - x) = - \sin x \\
 $
For $\cos $ function
$\cos x( - x) = \cos x$
$\cos ({180^{\circ}} - x) = - \cos x$
$\cos ({180^{\circ}} + x) = - \cos x$
$\cos ({90^{\circ}} + x) = - \sin x$
$\cos ({90^ \circ } - x) = \sin x$
$\cos ({360^{\circ}} - x) = \cos x$
Now, these are the results that would be used to prove the proof mentioned in this Question as using these identities, we would convert the left hand side that is L.H.S per the right hand side that is R.H.S to make either of them equal to the other.
In this particular question, we will first convert all the trigonometric functions in terms of $\sin $ and $\cos $ function and then we will try to make the L.H.S. and the R.H.S. equal.
As mentioned in the Question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows
\[
   = \dfrac{{\cos ({{180}^{\circ}} - A)\cot ({{90}^ \circ } + A)\cos ( - A)}}{{\tan ({{180}^{\circ}} + A)\tan ({{270}^{\circ}} + A)\sin ({{360}^{\circ}} - A)}} \\
   = \dfrac{{\cos ({{180}^{\circ}} - A)\dfrac{{\cos ({{90}^ \circ } + A)}}{{\sin ({{90}^ \circ } + A)}}\cos ( - A)}}{{\dfrac{{\sin ({{180}^{\circ}} + A)}}{{\cos ({{180}^{\circ}} + A)}}\dfrac{{\sin ({{270}^{\circ}} + A)}}{{\cos ({{270}^{\circ}} + A)}}\sin ({{360}^{\circ}} - A)}} \\
 \]
Now, on simplifying the angles of the trigonometric functions, we get the following result
$ = \dfrac{{( - \cos A)(\dfrac{{ - \sin A}}{{\cos A}})(\cos A)}}{{(\dfrac{{ - \sin A}}{{ - \cos A}})(\dfrac{{ - \cos A}}{{\sin A}})( - \sin A)}}$
$ = \cos A$

Note:
Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint.