Question

Find the value of \[\sin {{135}^{\circ }}\cos {{210}^{\circ }}\tan {{240}^{\circ }}\cot {{300}^{\circ }}\sec {{330}^{\circ }}\].

Answer 1

Hint: We can solve this with the help of quadrants. Take each value and find which quadrant it belongs to and simplify it. When all the simplified values are found use the trigonometric tables to get the value of the functions.



Complete step-by-step answer:
Let us use the ASTC rules to determine the sign of the ratio and quadrant expressions. Let us consider the original angle given as \[\theta \]and the auxiliary value as \[\alpha \], this can be generated according to the quadrant taken.

For quadrant II, we can write that, \[\theta =180-\alpha \].

For quadrant III, we can write that, \[\theta =180+\alpha \], and for quadrant IV, we can write that, \[\theta =360-\alpha \].

Now let us consider our \[{{1}^{st}}\]value of \[\sin {{135}^{\circ }}\]. \[{{135}^{\circ }}\]is in quadrant II.
So, \[\sin {{135}^{\circ }}=\sin \left( 180-\alpha \right)\],
i.e. \[\sin {{135}^{\circ }}=\sin \left( 180-{{45}^{\circ }} \right)\].
In quadrant II, sine is positive. Hence, \[\sin {{135}^{\circ }}=\sin \left( 180-{{45}^{\circ }} \right)=\sin {{45}^{\circ }}\].

Similarly, \[\cos {{210}^{\circ }}\]is in \[{{3}^{rd}}\]quadrant and cosine is negative in \[{{3}^{rd}}\]quadrant.
\[\cos {{210}^{\circ }}=\cos \left( 180+\alpha \right)=\cos \left( 180+{{30}^{\circ }} \right)\].
\[\therefore \cos {{210}^{\circ }}=\cos \left( 180+{{30}^{\circ }} \right)=-\cos {{30}^{\circ }}\] [As cosine is in \[{{3}^{rd}}\]quadrant, its negative]

Similarly, \[\tan {{240}^{\circ }}\]is in the \[{{3}^{rd}}\]quadrant, it is positive.
\[\tan {{240}^{\circ }}=\tan \left( 180+\alpha \right)=\tan \left( 180+60 \right)=\tan {{60}^{\circ }}\]

Now \[\cot {{300}^{\circ }}\]lies in the \[{{4}^{th}}\]quadrant, where it is negative.
\[\cot {{300}^{\circ }}=\cot \left( 360-\alpha \right)=\cot \left( 360-60 \right)=-\cot {{60}^{\circ }}\].

Similarly, \[\sec {{330}^{\circ }}\]lies in \[{{4}^{th}}\]quadrant, and it is positive.
\[\sec {{330}^{\circ }}=\sec \left( 360-\alpha \right)=\sec \left( 360-30 \right)=\sec {{30}^{\circ }}\]
Hence we can write,
\[\begin{align}
  & \sin {{135}^{\circ }}\cos {{210}^{\circ }}\tan {{240}^{\circ }}\cot {{300}^{\circ }}\sec {{330}^{\circ }} \\
 & =\sin {{45}^{\circ }}\left( -\cos {{30}^{\circ }} \right)\tan {{60}^{\circ }}\left( -\cot {{60}^{\circ }} \right)\sec {{30}^{\circ }} \\
 & =\sin {{45}^{\circ }}\cos {{30}^{\circ }}\tan {{60}^{\circ }}\cot {{60}^{\circ }}\sec {{30}^{\circ }} \\
\end{align}\]

From the trigonometric table we can find out the values.

	\[{{0}^{\circ }}\]	\[{{30}^{\circ }}\]	\[{{45}^{\circ }}\]	\[{{60}^{\circ }}\]	\[{{90}^{\circ }}\]
sin	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	1
cos	1	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	0
tan	0	\[\dfrac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	N.A
cosec	N.A	2	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	1
sec	1	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	N.A
cot	N.A	\[\sqrt{3}\]	1	\[\dfrac{1}{\sqrt{3}}\]	0

From this table we can say that,
\[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\]
\[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
\[\tan {{60}^{\circ }}=\sqrt{3}\]
\[\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}\]
\[\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}}\]
\[\begin{align}
  & \Rightarrow \sin {{45}^{\circ }}\cos {{30}^{\circ }}\tan {{60}^{\circ }}\cot {{60}^{\circ }}\sec {{30}^{\circ }} \\
 & =\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}\times \sqrt{3}\times \dfrac{1}{\sqrt{3}}\times \dfrac{2}{\sqrt{3}} \\
\end{align}\]
Cancel out like terms and simplify them, we get the final answer as,
\[\sin {{45}^{\circ }}\cos {{30}^{\circ }}\tan {{60}^{\circ }}\cot {{60}^{\circ }}\sec {{30}^{\circ }}=\dfrac{1}{\sqrt{2}}\].
Hence, we got \[\sin {{45}^{\circ }}\cos {{30}^{\circ }}\tan {{60}^{\circ }}\cot {{60}^{\circ }}\sec {{30}^{\circ }}=\dfrac{1}{\sqrt{2}}\].

Note: Remember the basics of quadrants that,
In quadrant 2: \[\sin \theta \ And \csc \theta \] are positive.
In quadrant 3: \[\tan \theta \ And \cot \theta \] are positive.
In quadrant 4: \[\cos \theta \ And \sec \theta \] are positive.
In quadrant 1: all are positive.