Question

Prove that:-\[\dfrac{\sin ({{180}^{\circ }}+\theta )\cdot \cos ({{90}^{\circ }}+\theta )\cdot \tan \left( {{270}^{\circ }}-\theta \right)\cdot \cot ({{360}^{\circ }}-\theta )}{\sin ({{360}^{\circ }}-\theta )\cdot \cos ({{360}^{\circ }}+\theta )\cdot \cos ec(-\theta )\cdot \sin \left( {{270}^{\circ }}+\theta \right)}=1\]

Answer 1

Hint: -In such question, 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 answer:
The below mentioned formulae may be used before solving, in the solution which is as follows
\[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \cot x=\dfrac{\cos x}{\sin x} \\
 & \cos ecx=\dfrac{1}{\sin x} \\
 & \sec x=\dfrac{1}{\cos x} \\
\end{align}\]
Some other important formulae that might be used in solving this question is as follows
\[\begin{align}
  & \sin (-x)=-\sin x \\
 & \sin ({{90}^{\circ }}-x)=\cos x \\
 & \sin ({{90}^{\circ }}+x)=\cos x \\
 & \sin ({{180}^{\circ }}-x)=\sin x \\
 & \sin ({{180}^{\circ }}+x)=-\sin x\ \ \ \ \ \\
 & \sin ({{270}^{\circ }}-x)=-\cos x \\
 & \sin ({{270}^{\circ }}+x)=-\cos x \\
 & \cos (-x)=\cos x \\
 & \cos ({{90}^{\circ }}-x)=\sin x \\
 & \cos ({{90}^{\circ }}+x)=-\sin x \\
 & \cos ({{180}^{\circ }}-x)=-\cos x \\
 & \cos ({{180}^{\circ }}+x)=-\cos x \\
 & \cos ({{270}^{\circ }}-x)=-\sin x \\
 & \cos ({{270}^{\circ }}+x)=\sin x \\
\end{align}\]
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. or 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
\[\begin{align}
  & =\dfrac{\sin ({{180}^{\circ }}+\theta )\cdot \cos ({{90}^{\circ }}+\theta )\cdot \tan \left( {{270}^{\circ }}-\theta \right)\cdot \cot ({{360}^{\circ }}-\theta )}{\sin ({{360}^{\circ }}-\theta )\cdot \cos ({{360}^{\circ }}+\theta )\cdot \cos ec(-\theta )\cdot \sin \left( {{270}^{\circ }}+\theta \right)} \\
 & =\dfrac{\sin ({{180}^{\circ }}+\theta )\cdot \cos ({{90}^{\circ }}+\theta )\cdot \dfrac{\sin \left( {{270}^{\circ }}-\theta \right)}{\cos \left( {{270}^{\circ }}-\theta \right)}\cdot \dfrac{\cos ({{360}^{\circ }}-\theta )}{\sin ({{360}^{\circ }}-\theta )}}{\sin ({{360}^{\circ }}-\theta )\cdot \cos ({{360}^{\circ }}+\theta )\cdot \dfrac{1}{\sin (-\theta )}\cdot \sin \left( {{270}^{\circ }}+\theta \right)} \\
\end{align}\]
Now, on simplifying the angles of the trigonometric functions, we get the following result
\[\begin{align}
  & =\dfrac{-\sin (\theta )\cdot -\sin (\theta )\cdot \dfrac{-\cos \left( \theta \right)}{-\sin \left( \theta \right)}\cdot \dfrac{\cos (\theta )}{-\sin (\theta )}}{-\sin (\theta )\cdot \cos (\theta )\cdot \dfrac{-1}{\sin (\theta )}\cdot -\cos \left( \theta \right)} \\
 & =\dfrac{-{{\cos }^{2}}\theta }{-{{\cos }^{2}}\theta } \\
\end{align}\]
(Using the identities that are mentioned in the hint)
Now, on cancelling the terms, we get the following
\[=1\]
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.

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. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.
Here, one more important thing is that
\[\left( {{360}^{\circ }}+x \right)=\left( x \right)\]
This is how the left hand angle can be treated as because the period of any trigonometric function is \[2\pi \] .