Question

Evaluate the following: \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-cosec{{60}^{\circ }}}{\sec {{30}^{\circ }}+cos{{60}^{\circ }}+\cot {{45}^{\circ }}}\].

Answer 1

Hint: We will use the trigonometric ratio table and find out the values of trigonometric ratios as follows:
\[\sin {{30}^{\circ }}=\dfrac{1}{2},tan{{45}^{\circ }}=1,cosec{{60}^{\circ }}=\dfrac{2}{\sqrt{3}},\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}},\cos {{60}^{\circ }}=\dfrac{1}{2},\cot {{45}^{\circ }}=1\]
Then we will substitute the values of those trigonometric ratios in the equation given in the equation and simplify to get the answer.

Complete step-by-step answer:
We have been given the expression \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-cosec{{60}^{\circ }}}{\sec {{30}^{\circ }}+cos{{60}^{\circ }}+\cot {{45}^{\circ }}}\].
We can refer to the following table for finding the value of trigonometric functions at standard angles.

Ratio/Angle\[\theta \]	\[{{0}^{\circ }}\]	\[{{30}^{\circ }}\]	\[{{45}^{\circ }}\]	\[{{60}^{\circ }}\]	\[{{90}^{\circ }}\]
\[\sin \theta \]	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	1
\[\cos \theta \]	1	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	0
\[tan\theta \]	0	\[\dfrac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	Not defined
\[cosec\theta \]	Not defined	2	\[\sqrt{2}\]	\[\dfrac{2}{\sqrt{3}}\]	1
\[\sec \theta \]	1	\[\dfrac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	Not defined
\[\cot \theta \]	Not defined	\[\sqrt{3}\]	1	\[\dfrac{1}{\sqrt{3}}\]	0

Referring to the table above, we can see that \[\sin {{30}^{\circ }}=\dfrac{1}{2},tan{{45}^{\circ }}=1,cosec{{60}^{\circ }}=\dfrac{2}{\sqrt{3}},\sec {{30}^{\circ }}=\dfrac{2}{\sqrt{3}},\cos {{60}^{\circ }}=\dfrac{1}{2},\cot {{45}^{\circ }}=1\].
So by substituting these values in the equation given in the question, we get as follows:
\[ \Rightarrow \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-cosec{{60}^{\circ }}}{\sec {{30}^{\circ }}+cos{{60}^{\circ }}+\cot {{45}^{\circ }}} \]
\[ \Rightarrow \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1} \]
\[ \Rightarrow \dfrac{\dfrac{1+2}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1+2}{2}} \]
\[ \Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{3}{2}} \]
On multiplying the numerator as well as the denominator by \[\left( \dfrac{3}{2}-\dfrac{2}{\sqrt{3}} \right)\] in order to rationalize the denominator, we get as follows:
\[ \Rightarrow \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-cosec{{60}^{\circ }}}{\sec {{30}^{\circ }}+cos{{60}^{\circ }}+\cot {{45}^{\circ }}} \]

\[ \Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}\times \left( \dfrac{3}{2}-\dfrac{2}{\sqrt{3}} \right)}{\dfrac{3}{2}+\dfrac{2}{\sqrt{3}}\times \left( \dfrac{3}{2}-\dfrac{2}{\sqrt{3}} \right)} \]
Since we know the trigonometric identity \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\], we apply it in the equation and get as follows:
\[\Rightarrow \dfrac{{{\left( \dfrac{3}{2}-\dfrac{2}{\sqrt{3}} \right)}^{2}}}{{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}}\]
\[\Rightarrow \dfrac{{{\left( \dfrac{3}{2} \right)}^{2}}+{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-2\times \dfrac{3}{2}\times \dfrac{2}{\sqrt{3}}}{\dfrac{9}{4}-\dfrac{4}{3}}\]
On taking LCM of 3 and 4, we get as follows:
\[ \Rightarrow \dfrac{\dfrac{9}{4}+\dfrac{4}{3}-\dfrac{6}{\sqrt{3}}}{\dfrac{27-16}{12}} \]
\[ \Rightarrow \dfrac{\dfrac{27+16}{12}-\dfrac{6\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}}{\dfrac{11}{12}} \]
\[ \Rightarrow \dfrac{\dfrac{43}{12}-\dfrac{6\sqrt{3}}{3}}{\dfrac{11}{12}} \]
Taking LCM of 12 and 3, we get as follows:
\[ \Rightarrow \dfrac{\dfrac{43-24\sqrt{3}}{12}}{\dfrac{11}{12}} \]
\[ \Rightarrow \dfrac{\left( 43-24\sqrt{3} \right)\times 12}{12\times 11} \]
\[ \Rightarrow \dfrac{43-24\sqrt{3}}{11} \]
Therefore, the required value of the given expression is \[\dfrac{43-24\sqrt{3}}{11}\].

Note: Don’t get confused while substituting the value of trigonometric ratios in the given expression. Also, be careful while calculating and take care of the signs. In this type of question, you must have to remember the trigonometric ratio table so that we can easily substitute any values. We can convert all the values in terms of sin and cos because so many students remember only these formulas.