Question

Find the value of \[\dfrac{\cot {{54}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}\].

Answer 1

Hint: In this question, first identify the complementary angle. Now use \[\tan \theta =\cot \left( 90-\theta \right)\] and \[\cot \theta =\tan \left( 90-\theta \right)\] to equate the numerator and denominator in terms. Finally, simplify the expression to get the required value.

Complete step-by-step answer:
In this question, we have to find the value of \[\dfrac{\cot {{54}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}\]. Before proceeding with this question, let us first understand what complementary angles are. Complementary angles are angles whose sum is equal to \[{{90}^{o}}\]. If we have, \[\angle A+\angle B={{90}^{o}}\], then \[\angle A\] and \[\angle B\] are complementary angles of each other.
Similarly, \[\theta \] and \[\left( {{90}^{o}}-\theta \right)\] are complementary to each other because \[\theta +{{90}^{o}}-\theta ={{90}^{o}}\].
So, in trigonometry, we have multiple formulas related to complementary angles and that are the following:
\[\begin{align}
  & \sin \left( 90-\theta \right)=\cos \theta ;\cos \left( 90-\theta \right)=\sin \theta \\
 & \cot \left( 90-\theta \right)=\tan \theta ;\tan \left( 90-\theta \right)=\cot \theta \\
 & \operatorname{cosec}\left( 90-\theta \right)=\sec \theta ;\sec \left( 90-\theta \right)=\operatorname{cosec}\theta \\
\end{align}\]
Let us now consider the expression given in the question,
\[E=\dfrac{\cot {{54}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}....\left( i \right)\]
In the above expression, we can see that,
\[{{54}^{o}}+{{36}^{o}}={{90}^{o}}\]
So, these angles are complementary. We know that \[\cot \theta =\tan \left( 90-\theta \right)\]. So, by substituting \[\theta ={{54}^{o}}\], we get,
\[\cot {{54}^{o}}=\tan \left( {{90}^{o}}-{{54}^{o}} \right)=\tan {{36}^{o}}\]
So, by substituting the value of \[\cot {{54}^{o}}\] in the above expression, we get,
\[E=\dfrac{\tan {{36}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}\]
Similarly, we know that \[{{20}^{o}}+{{70}^{o}}={{90}^{o}}\]. So, these angles are also complementary.
We know that \[\cot \theta =\tan \left( {{90}^{o}}-\theta \right)\]. So, by substituting \[\theta ={{70}^{o}}\], we get,
\[\cot {{70}^{o}}=\tan \left( {{90}^{o}}-{{70}^{o}} \right)=\tan {{20}^{o}}\]
So by substituting the value of \[\cot {{70}^{o}}\] in the above expression, we get,
\[E=\dfrac{\tan {{36}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\tan {{20}^{o}}}\]
Now, by canceling the like terms in the above expression, we get,
\[E=1+1=2\]
So, we have got the value of \[\dfrac{\cot {{54}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}=2\]

Note: Students can also solve this question in the following way
\[E=\dfrac{\cot {{54}^{o}}}{\tan {{36}^{o}}}+\dfrac{\tan {{20}^{o}}}{\cot {{70}^{o}}}\]
We know that \[\tan \theta =\cot \left( 90-\theta \right)\]. So by using this, we get,
\[E=\dfrac{\cot {{54}^{o}}}{\cot \left( 90-{{36}^{o}} \right)}+\dfrac{\cot \left( {{90}^{o}}-{{20}^{o}} \right)}{\cot {{70}^{o}}}\]
\[E=\dfrac{\cot {{54}^{o}}}{\cot {{54}^{o}}}+\dfrac{\cot {{70}^{o}}}{\cot {{70}^{o}}}\]
E = 1 + 1
E = 2