Question

What trigonometric ratio of angles from ${{0}^{\circ }}\text{ to }{{90}^{\circ }}$ are not defined?

Answer 1

Hint: For solving this sum, we need to draw the trigonometric ratio table for angles ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}\text{ and }{{90}^{\circ }}$. Then from the trigonometric table we can identify the ratios which are not defined. For drawing table, we need to memorize value of $\sin \theta $ only other can be calculated using following formula:
\[\begin{align}
  & \left( i \right)\cos \theta =\sin \left( {{90}^{\circ }}-\theta \right) \\
 & \left( ii \right)\tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \left( iii \right)\cot \theta =\dfrac{1}{\tan \theta } \\
 & \left( iv \right)\text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \left( v \right)\sec \theta =\dfrac{1}{\cos \theta } \\
\end{align}\]

Complete step-by-step answer:
Here we need to draw the trigonometric ratio table for angles ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}\text{ and }{{90}^{\circ }}$.
For the drawing table, let us first find values for the table. For $\sin \theta $ we just need to memorize the value. Values of $\sin \theta $ are:
$\sin {{0}^{\circ }}=0,\sin {{30}^{\circ }}=\dfrac{1}{2},\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\text{ and }\sin {{90}^{\circ }}=1$.
For finding values of $\cos \theta $ we will just use the formula $\cos \theta =\sin \left( {{90}^{\circ }}-\theta \right)$. So, the value of $\cos \theta =\sin \left( {{90}^{\circ }}-\theta \right)=\sin {{90}^{\circ }}=1$.
Similarly,
\[\begin{align}
  & \Rightarrow \cos {{30}^{\circ }}=\sin \left( {{90}^{\circ }}-{{30}^{\circ }} \right)=\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \cos {{45}^{\circ }}=\sin \left( {{90}^{\circ }}-{{45}^{\circ }} \right)=\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\
 & \Rightarrow \cos {{60}^{\circ }}=\sin \left( {{90}^{\circ }}-{{60}^{\circ }} \right)=\sin {{30}^{\circ }}=\dfrac{1}{2} \\
 & \Rightarrow \cos {{90}^{\circ }}=\sin \left( {{90}^{\circ }}-{{90}^{\circ }} \right)=\sin {{0}^{\circ }}=0 \\
\end{align}\]
So value of $\cos \theta $ become as:
$\cos {{0}^{\circ }}=1,\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{60}^{\circ }}=\dfrac{1}{2},\cos {{90}^{\circ }}=0$.
Now let us find value of $\tan \theta $.
As we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ so we get value of
\[\begin{align}
  & \Rightarrow \tan {{0}^{\circ }}=\dfrac{\sin {{0}^{\circ }}}{\cos {{0}^{\circ }}}=\dfrac{0}{1}=0 \\
 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{\sin {{30}^{\circ }}}{\cos {{30}^{\circ }}}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{1}{2}\times \dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow \tan {{45}^{\circ }}=\dfrac{\sin {{45}^{\circ }}}{\cos {{45}^{\circ }}}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}}=1 \\
 & \Rightarrow \tan {{60}^{\circ }}=\dfrac{\sin {{60}^{\circ }}}{\cos {{60}^{\circ }}}=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}=\dfrac{\sqrt{3}}{2}\times \dfrac{2}{1}=\sqrt{3} \\
 & \Rightarrow \tan {{90}^{\circ }}=\dfrac{\sin {{90}^{\circ }}}{\cos {{90}^{\circ }}}=\dfrac{1}{0}=\text{not defined} \\
\end{align}\]
Hence the value of $\tan \theta $ are:
$\tan {{0}^{\circ }}=0,\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}},\tan {{45}^{\circ }}=1,\tan {{60}^{\circ }}=\sqrt{3},\tan {{90}^{\circ }}=\text{not defined}$
Now we need to find the values of $\cot \theta $. As we know $\cot \theta =\dfrac{1}{\tan \theta }$ so the value of $\cot \theta $ will just be reciprocal of $\tan \theta $. Hence, we can find value as:
\[\begin{align}
  & \Rightarrow \cot {{0}^{\circ }}=\dfrac{1}{\tan {{0}^{\circ }}}=\dfrac{1}{0}=\text{not defined} \\
 & \Rightarrow \cot {{30}^{\circ }}=\dfrac{1}{\tan {{30}^{\circ }}}=\dfrac{1}{\dfrac{1}{\sqrt{3}}}=\sqrt{3} \\
 & \Rightarrow \cot {{45}^{\circ }}=\dfrac{1}{\tan {{45}^{\circ }}}=\dfrac{1}{1}=1 \\
 & \Rightarrow \cot {{60}^{\circ }}=\dfrac{1}{\tan {{60}^{\circ }}}=\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow \cot {{90}^{\circ }}=\dfrac{1}{\tan {{90}^{\circ }}}=\dfrac{1}{\dfrac{1}{0}}=\text{0} \\
\end{align}\]
Hence we have found values of $\cot \theta $ also.
Now for finding the value of $\text{cosec}\theta $ we just need to take reciprocal of values of $\sin \theta $ as $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ so we get:
\[\begin{align}
  & \Rightarrow \text{cosec}{{0}^{\circ }}=\dfrac{1}{\sin {{0}^{\circ }}}=\dfrac{1}{0}=\text{not defined} \\
 & \Rightarrow \text{cosec}{{30}^{\circ }}=\dfrac{1}{\sin {{30}^{\circ }}}=\dfrac{1}{\dfrac{1}{2}}=2 \\
 & \Rightarrow \text{cosec}{{45}^{\circ }}=\dfrac{1}{\sin {{45}^{\circ }}}=\dfrac{1}{\dfrac{1}{\sqrt{2}}}=\sqrt{2} \\
 & \Rightarrow \text{cosec}{{60}^{\circ }}=\dfrac{1}{\sin {{60}^{\circ }}}=\dfrac{1}{\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}} \\
 & \Rightarrow \text{cosec}{{90}^{\circ }}=\dfrac{1}{\sin {{90}^{\circ }}}=\dfrac{1}{1}=\text{1} \\
\end{align}\]
Hence we have found values of $\text{cosec}\theta $ also.
Now, for finding values of $\sec \theta $ we just need to take reciprocal of $\cos \theta $ as $\sec \theta =\dfrac{1}{\cos \theta }$s so we get:
\[\begin{align}
  & \Rightarrow \text{sec}{{0}^{\circ }}=\dfrac{1}{\cos {{0}^{\circ }}}=\dfrac{1}{1}=\text{1} \\
 & \Rightarrow \text{sec}{{30}^{\circ }}=\dfrac{1}{\cos {{30}^{\circ }}}=\dfrac{1}{\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}} \\
 & \Rightarrow \text{sec}{{45}^{\circ }}=\dfrac{1}{\cos {{45}^{\circ }}}=\dfrac{1}{\dfrac{1}{\sqrt{2}}}=\sqrt{2} \\
 & \Rightarrow \text{sec}{{60}^{\circ }}=\dfrac{1}{\cos {{60}^{\circ }}}=\dfrac{1}{\dfrac{1}{2}}=2 \\
 & \Rightarrow \text{sec}{{90}^{\circ }}=\dfrac{1}{\cos {{90}^{\circ }}}=\dfrac{1}{0}=\text{not defined} \\
\end{align}\]
Hence we have found values of $\sec \theta $ also.
Putting all these values in the form of table, we get trigonometric ratio table as:

Angles	${{0}^{\circ }}$	${{30}^{\circ }}$	${{45}^{\circ }}$	${{60}^{\circ }}$	\[{{90}^{\circ }}\]
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	$\sqrt{3}$	$\infty $
cot	$\infty $	$\sqrt{3}$	1	$\dfrac{1}{\sqrt{3}}$	0
cosec	$\infty $	2	$\sqrt{2}$	$\dfrac{2}{\sqrt{3}}$	1
sec	1	$\dfrac{2}{\sqrt{3}}$	$\sqrt{2}$	2	$\infty $

From the table we can find values that are not defined which are:
$\tan {{90}^{\circ }},\cot {{0}^{\circ }},\text{cosec}{{0}^{\circ }}\text{ and }\sec {{90}^{\circ }}$.
Hence required trigonometric ratios are $\tan {{90}^{\circ }},\cot {{0}^{\circ }},\text{cosec}{{0}^{\circ }}\text{ and }\sec {{90}^{\circ }}$.

So, the correct answer is “Option A”.

Note: Students can memorize the values of $\sin \theta $ in following way:
Divide 0, 1, 2, 3, 4 by 4 and take under root. Then values of $\sin \theta $ will be:

sin${{0}^{\circ }}$	sin${{30}^{\circ }}$	sin${{45}^{\circ }}$	sin${{60}^{\circ }}$	sin\[{{90}^{\circ }}\]
$\sqrt{\dfrac{0}{4}=0}$	$\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$	$\sqrt{\dfrac{2}{4}}=\dfrac{1}{\sqrt{2}}$	$\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}$	$\sqrt{\dfrac{4}{4}}=1$

Students should note that, ratios whose values are $\dfrac{1}{0},\infty $ are considered as not defined values. Take care while finding the values.