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:

& \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{aligned}$$ Let us now consider the expression given in the question, $$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