Question

Find the value of $\dfrac{\sin {{30}^{0}}+\tan {{45}^{0}}-\csc {{60}^{0}}}{\cot {{45}^{0}}+\cos {{60}^{0}}-\sec {{30}^{0}}}$.

Answer 1

Hint: For solving this as we can see that in the given expression there are some standard trigonometric ratios. So, we will put the value of each term in the given expression directly and then solve for the correct answer.

Complete step-by-step answer:
Given:
We have to evaluate the value of $\dfrac{\sin {{30}^{0}}+\tan {{45}^{0}}-\csc {{60}^{0}}}{\cot {{45}^{0}}+\cos {{60}^{0}}-\sec {{30}^{0}}}$.
Now, before we proceed first we should know the following results:
$\begin{align}
  & \sin {{30}^{0}}=\dfrac{1}{2}.............\left( 1 \right) \\
 & \cos {{60}^{0}}=\dfrac{1}{2}.............\left( 2 \right) \\
 & \cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}...........\left( 3 \right) \\
 & \sin {{60}^{0}}=\dfrac{\sqrt{3}}{2}............\left( 4 \right) \\
 & \tan {{45}^{0}}=1................\left( 5 \right) \\
 & \csc \theta =\dfrac{1}{\sin \theta }.............\left( 6 \right) \\
 & \sec \theta =\dfrac{1}{\cos \theta }.............\left( 7 \right) \\
 & \cot \theta =\dfrac{1}{\tan \theta }..............\left( 8 \right) \\
\end{align}$
Now, from the above results, we can easily solve this question. Using equation (5) and equation (3) to find the value of $\csc {{60}^{0}}$ . Then,
$\begin{align}
  & \csc \theta =\dfrac{1}{\sin \theta } \\
 & \Rightarrow \csc {{60}^{0}}=\dfrac{1}{\sin {{60}^{0}}} \\
 & \Rightarrow \csc {{60}^{0}}=\dfrac{2}{\sqrt{3}}................\left( 9 \right) \\
\end{align}$
Now, using equation (4) and equation (7) to find the value of $\cot {{45}^{0}}$ . Then,
$\begin{align}
  & \cot \theta =\dfrac{1}{\tan \theta } \\
 & \Rightarrow \cot {{45}^{0}}=\dfrac{1}{\tan {{45}^{0}}} \\
 & \Rightarrow \cot {{45}^{0}}=1..................\left( 10 \right) \\
\end{align}$
Now, using equation (3) and equation (7) to find the value of $\sec {{30}^{0}}$ . Then,
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \Rightarrow \sec {{30}^{0}}=\dfrac{1}{\cos {{30}^{0}}} \\
 & \Rightarrow \sec {{30}^{0}}=\dfrac{2}{\sqrt{3}}................\left( 11 \right) \\
\end{align}$
Now, we will directly put the value of each term in $\dfrac{\sin {{30}^{0}}+\tan {{45}^{0}}-\csc {{60}^{0}}}{\cot {{45}^{0}}+\cos {{60}^{0}}-\sec {{30}^{0}}}$ from the equation (1), equation (2), equation (5), equation (9), equation (10) and equation (11). Then,
$\begin{align}
  & \dfrac{\sin {{30}^{0}}+\tan {{45}^{0}}-\csc {{60}^{0}}}{\cot {{45}^{0}}+\cos {{60}^{0}}-\sec {{30}^{0}}} \\
 & \Rightarrow \left( \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{2}-\dfrac{2}{\sqrt{3}}} \right)=\left( \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}} \right) \\
 & \Rightarrow 1 \\
\end{align}$
Thus, from the above calculation, we can say that $\dfrac{\sin {{30}^{0}}+\tan {{45}^{0}}-\csc {{60}^{0}}}{\cot {{45}^{0}}+\cos {{60}^{0}}-\sec {{30}^{0}}}$ is equal to 1.

Hint: The question was very easy to solve if we know the values of each term then by avoiding calculation mistakes we can get the correct answer. Moreover, there is another short by which one can directly answer even if we don’t know the values. If $\alpha $ and $\beta $ are two angles such that, $\alpha +\beta ={{90}^{0}}$. Then, $\sin \alpha =\cos \beta $, $\tan \alpha =\cot \beta $ and $\sec \alpha =\csc \beta $. When we put the proper values of $\alpha $ and $\beta $ then we can analyse whether the numerator and denominator of the given expression is equal.