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Question: Solve: \(\dfrac{\sin \left( {{180}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}+\theta \rig...

Solve:
sin⁡(180∘+θ)cos⁡(90∘+θ)tan⁡(270∘−θ)cot⁡(360∘+θ)sin⁡(360∘+θ)cos⁡(360∘+θ)csc⁡(−θ)sin⁡(270∘+θ)\dfrac{\sin \left( {{180}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}+\theta \right)\tan \left( {{270}^{\circ }}-\theta \right)\cot \left( {{360}^{\circ }}+\theta \right)}{\sin \left( {{360}^{\circ }}+\theta \right)\cos \left( {{360}^{\circ }}+\theta \right)\csc \left( -\theta \right)\sin \left( {{270}^{\circ }}+\theta \right)}
(a) 2\sqrt{2}
(b) 32\dfrac{\sqrt{3}}{2}
(c) 32\dfrac{3}{2}
(d) 1

Explanation

Solution

To solve the above expression, we are going to use the following trigonometric angle conversions:
sin⁡(180∘+θ)=−sin⁡θ\sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta , cos⁡(90∘+θ)=−sin⁡θ\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta , tan⁡(270∘−θ)=cot⁡θ\tan \left( {{270}^{\circ }}-\theta \right)=\cot \theta , cot⁡(360∘+θ)=cot⁡θ\cot \left( {{360}^{\circ }}+\theta \right)=\cot \theta , sin⁡(360∘+θ)=sin⁡θ\sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta , cos⁡(360∘+θ)=cos⁡θ\cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta , csc⁡(−θ)=−csc⁡θ\csc \left( -\theta \right)=-\csc \theta and sin⁡(270∘+θ)=−cos⁡θ\sin \left( {{270}^{\circ }}+\theta \right)=-\cos \theta . Now, substitute these conversions in the given trigonometric expression and then solve that expression. While solving that expression, we need to use the following trigonometric properties also: cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta }. Hence, this is how we are going to solve the given expression.

Complete step-by-step answer:
The trigonometric expression given in the above problem which we are going to solve is as follows:
sin⁡(180∘+θ)cos⁡(90∘+θ)tan⁡(270∘−θ)cot⁡(360∘+θ)sin⁡(360∘+θ)cos⁡(360∘+θ)csc⁡(−θ)sin⁡(270∘+θ)\dfrac{\sin \left( {{180}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}+\theta \right)\tan \left( {{270}^{\circ }}-\theta \right)\cot \left( {{360}^{\circ }}+\theta \right)}{\sin \left( {{360}^{\circ }}+\theta \right)\cos \left( {{360}^{\circ }}+\theta \right)\csc \left( -\theta \right)\sin \left( {{270}^{\circ }}+\theta \right)}
To solve the above trigonometric expressions we need to have knowledge of trigonometric angle conversions which we are going to show in the below:
sin⁡(180∘+θ)=−sin⁡θ; cos⁡(90∘+θ)=−sin⁡θ; tan⁡(270∘−θ)=cot⁡θ; cot⁡(360∘+θ)=cot⁡θ; sin⁡(360∘+θ)=sin⁡θ; cos⁡(360∘+θ)=cos⁡θ; csc⁡(−θ)=−csc⁡θ; sin⁡(270∘+θ)=−cos⁡θ \begin{aligned} & \sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta ; \\\ & \cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta ; \\\ & \tan \left( {{270}^{\circ }}-\theta \right)=\cot \theta ; \\\ & \cot \left( {{360}^{\circ }}+\theta \right)=\cot \theta ; \\\ & \sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta ; \\\ & \cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta ; \\\ & \csc \left( -\theta \right)=-\csc \theta ; \\\ & \sin \left( {{270}^{\circ }}+\theta \right)=-\cos \theta \\\ \end{aligned}
Now, we are going to substitute the above trigonometric angle conversions in the given expression and we get,
−sin⁡θ(−sin⁡θ)cot⁡θcot⁡θsin⁡θcos⁡θ(−csc⁡θ)(−cos⁡θ)\dfrac{-\sin \theta \left( -\sin \theta \right)\cot \theta \cot \theta }{\sin \theta \cos \theta \left( -\csc \theta \right)\left( -\cos \theta \right)}
We know that when we multiply two negative signs then we get positive sign so the two negative signs multiplying in the numerator and the denominator we get,
sin⁡θ(sin⁡θ)cot⁡θcot⁡θsin⁡θcos⁡θ(csc⁡θ)(cos⁡θ)\dfrac{\sin \theta \left( \sin \theta \right)\cot \theta \cot \theta }{\sin \theta \cos \theta \left( \csc \theta \right)\left( \cos \theta \right)} …………. (1)
Now, we are going to use the following trigonometric conversions of the trigonometric ratios.
cot⁡θ=cos⁡θsin⁡θ; csc⁡θ=1sin⁡θ \begin{aligned} & \cot \theta =\dfrac{\cos \theta }{\sin \theta }; \\\ & \csc \theta =\dfrac{1}{\sin \theta } \\\ \end{aligned}
Using the above trigonometric relations in (1) we get,
sin⁡θ(sin⁡θ)(cos⁡θsin⁡θ)(cos⁡θsin⁡θ)sin⁡θcos⁡θ(1sin⁡θ)(cos⁡θ)\dfrac{\sin \theta \left( \sin \theta \right)\left( \dfrac{\cos \theta }{\sin \theta } \right)\left( \dfrac{\cos \theta }{\sin \theta } \right)}{\sin \theta \cos \theta \left( \dfrac{1}{\sin \theta } \right)\left( \cos \theta \right)}
In the numerator of the above expression, sin⁡θ(sin⁡θ)\sin \theta \left( \sin \theta \right) will be cancelled out and in the denominator of the above expression, sin⁡θ\sin \theta will get cancelled out. Then the above expression will look like:
cos⁡θ(cos⁡θ)cos⁡θ(cos⁡θ)\dfrac{\cos \theta \left( \cos \theta \right)}{\cos \theta \left( \cos \theta \right)}
Now, numerator and denominator is same so both these terms written in the numerator and the denominator will get cancelled out and we get,
1
Hence, the solution of the given expression is 1 and the correct option is 1.

So, the correct answer is “Option (a)”.

Note: The trick to remember the trigonometric conversions which we have done above is that in the first quadrant all sin⁡,cos⁡,tan⁡,csc⁡,sec⁡,cot⁡\sin ,\cos ,\tan ,\csc ,\sec ,\cot are positive. In the second quadrant, sin⁡,csc⁡\sin ,\csc is positive, in the third quadrant tan⁡,cot⁡\tan ,\cot is positive and in the fourth quadrant cos⁡,sec⁡\cos ,\sec .