Question

In triangle $\Delta ABC$ , if $a=5$ , $b=4$ , $A=\dfrac{\pi }{2}+B$ then $\tan C=$
(A) $\dfrac{9}{40}$
(B) $4$
(C) $2.5$
(D) $1.45$

Answer 1

Hint: In this question we have been asked to find the value of $\tan C$ when in $\Delta ABC$ $a=5$ , $b=4$ and $A=\dfrac{\pi }{2}+B$ . For doing that we will use the Napier’s trigonometric formula given as follows $\tan \dfrac{A-B}{2}=\dfrac{a-b}{a+b}\cot \dfrac{C}{2}$ and the trigonometric formula $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ .

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $\tan C$ when in $\Delta ABC$ $a=5$ , $b=4$ and $A=\dfrac{\pi }{2}+B$ .

From the basic concepts of trigonometry we have learnt the Napier’s trigonometric formula given as follows $\tan \dfrac{A-B}{2}=\dfrac{a-b}{a+b}\cot \dfrac{C}{2}$ , $\tan \dfrac{B-C}{2}=\dfrac{b-c}{b+c}\cot \dfrac{A}{2}$ and $\tan \dfrac{C-A}{2}=\dfrac{c-a}{c+a}\cot \dfrac{B}{2}$ .
By using this we will have
 $\begin{align}
  & \Rightarrow \tan \dfrac{\left( \dfrac{\pi }{2} \right)}{2}=\dfrac{5-4}{5+4}\cot \dfrac{C}{2} \\
 & \Rightarrow \tan \dfrac{\pi }{4}=\dfrac{1}{9}\cot \dfrac{C}{2} \\
 & \Rightarrow 9=\cot \dfrac{C}{2} \\
\end{align}$ .
Hence we can say that $\tan \dfrac{C}{2}=\dfrac{1}{9}$ .
From the basic concepts of trigonometry we have learnt the trigonometric formula given as $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ .
By applying the above formula we will have
$\begin{align}
  & \Rightarrow \tan 2\left( \dfrac{C}{2} \right)=\dfrac{2\tan \left( \dfrac{C}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{C}{2} \right)} \\
 & \Rightarrow \tan C=\dfrac{2\left( \dfrac{1}{9} \right)}{1-{{\left( \dfrac{1}{9} \right)}^{2}}} \\
\end{align}$
By further simplifying this we will have
$\begin{align}
  & \Rightarrow \tan C=\dfrac{\left( \dfrac{2}{9} \right)}{\dfrac{80}{81}} \\
 & \Rightarrow \tan C=\dfrac{2\times 9}{80} \\
 & \Rightarrow \tan C=\dfrac{9}{40} \\
\end{align}$
Hence we can conclude that the value of $\tan C$ will be given as $\dfrac{9}{40}$ when in $\Delta ABC$ $a=5$ ,$b=4$ and $A=\dfrac{\pi }{2}+B$ .
Hence we can mark the option “A” as correct.

Note: While answering questions of this type we should be sure with our concepts that we are going to apply and the calculations that we are going to perform in between the steps. Someone can unintentionally make a calculation mistake in between the steps which will lead them to end up having a wrong conclusion. For example if we consider

$\begin{align}
  & \Rightarrow \tan 2\left( \dfrac{C}{2} \right)=\dfrac{2\tan \left( \dfrac{C}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{C}{2} \right)} \\
 & \Rightarrow \tan C=\dfrac{2\left( 9 \right)}{1-{{\left( 9 \right)}^{2}}} \\
 & \Rightarrow \tan C=\dfrac{\left( 18 \right)}{-80} \\
 & \Rightarrow \tan C=\dfrac{-9}{40} \\
\end{align}$
which is a wrong conclusion clearly. So we should be careful with the calculations and concepts.