Question

The value of the angle $ {{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ in degrees is equal to
(a) $ -{{20}^{\circ }} $
(b) $ {{20}^{\circ }} $
(c) $ {{25}^{\circ }} $
(d) $ {{40}^{\circ }} $

Answer 1

Hint: We start solving the problem by assigning the variable for $ {{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ . We then make use of the fact $ \left( \tan \left( A-B \right) \right)\times \left( 1+\tan A\tan B \right)=\tan A-\tan B $ by assuming suitable values for angles A and B. We then make the necessary calculations and make use of the facts $ \tan \left( 90-\theta \right)=\cot \theta $ and $ \tan \theta \times \cot \theta =1 $ to proceed through the problem. We the make the necessary calculations and use the fact $ {{\tan }^{-1}}\left( \tan \theta \right)=\theta $ , if $ \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right) $ to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of angle $ {{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ in degrees.
Let us assume $ \alpha ={{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ .
 $ \Rightarrow \tan \alpha =\tan {{65}^{\circ }}-2\tan {{40}^{\circ }} $ .
 $ \Rightarrow \tan \alpha =\tan {{65}^{\circ }}-\tan {{40}^{\circ }}-\tan {{40}^{\circ }} $ ---(1).
We know that $ \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} $ .
 $ \Rightarrow \left( \tan \left( A-B \right) \right)\times \left( 1+\tan A\tan B \right)=\tan A-\tan B $ . Let us make use of this result in equation (1) by assuming $ A={{65}^{\circ }} $ and $ B={{40}^{\circ }} $ .
\[\Rightarrow \tan \alpha =\left( \tan \left( {{65}^{\circ }}-{{40}^{\circ }} \right) \right)\times \left( 1+\tan {{65}^{\circ }}\tan {{40}^{\circ }} \right)-\tan {{40}^{\circ }}\].
\[\Rightarrow \tan \alpha =\left( \tan {{25}^{\circ }} \right)\times \left( 1+\tan {{65}^{\circ }}\tan {{40}^{\circ }} \right)-\tan {{40}^{\circ }}\].
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}+\tan {{25}^{\circ }}\tan {{65}^{\circ }}\tan {{40}^{\circ }}-\tan {{40}^{\circ }}\].
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}+\tan \left( {{90}^{\circ }}-{{65}^{\circ }} \right)\tan {{65}^{\circ }}\tan {{40}^{\circ }}-\tan {{40}^{\circ }}\].
We know that $ \tan \left( 90-\theta \right)=\cot \theta $ .
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}+\cot {{65}^{\circ }}\tan {{65}^{\circ }}\tan {{40}^{\circ }}-\tan {{40}^{\circ }}\].
We know that $ \tan \theta \times \cot \theta =1 $ .
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}+\left( 1\times \tan {{40}^{\circ }} \right)-\tan {{40}^{\circ }}\].
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}+\tan {{40}^{\circ }}-\tan {{40}^{\circ }}\].
\[\Rightarrow \tan \alpha =\tan {{25}^{\circ }}\].
\[\Rightarrow \alpha ={{\tan }^{-1}}\left( \tan {{25}^{\circ }} \right)\].
We know that $ {{\tan }^{-1}}\left( \tan \theta \right)=\theta $ , if $ \theta \in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right) $ .
\[\Rightarrow \alpha ={{25}^{\circ }}\].
But we already have $ \alpha ={{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ .
 $ \Rightarrow {{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right)={{25}^{\circ }} $ .
So, we have found the value of $ {{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ as $ {{25}^{\circ }} $ .
 $ \therefore, $ The correct option for the given problem is (c).

Note:
 We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also solve the given problem as shown below:
We have $ \alpha ={{\tan }^{-1}}\left( \tan {{65}^{\circ }}-2\tan {{40}^{\circ }} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \tan {{65}^{\circ }}-\tan {{40}^{\circ }}-\tan {{40}^{\circ }} \right) $ ---(2).
Now, let us solve $ \tan {{65}^{\circ }}-\tan {{40}^{\circ }} $ . We know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ .
 $ \Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin {{65}^{\circ }}}{\cos {{65}^{\circ }}}-\dfrac{\sin {{40}^{\circ }}}{\cos {{40}^{\circ }}} $ .
 $ \Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin {{65}^{\circ }}\cos {{40}^{\circ }}-\sin {{40}^{\circ }}\cos {{65}^{\circ }}}{\cos {{65}^{\circ }}\cos {{40}^{\circ }}} $ .
We know that $ \sin A\cos B-\cos A\sin B=\sin \left( A-B \right) $ .
 $ \Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin \left( {{65}^{\circ }}-{{40}^{\circ }} \right)}{\cos {{65}^{\circ }}\cos {{40}^{\circ }}} $ .
 $ \Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin {{25}^{\circ }}}{\cos {{65}^{\circ }}\cos {{40}^{\circ }}} $ .
\[\Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin {{25}^{\circ }}}{\cos \left( {{90}^{\circ }}-{{25}^{\circ }} \right)\cos {{40}^{\circ }}}\].
We know that $ \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ .
\[\Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{\sin {{25}^{\circ }}}{\sin {{25}^{\circ }}\cos {{40}^{\circ }}}\].
\[\Rightarrow \tan {{65}^{\circ }}-\tan {{40}^{\circ }}=\dfrac{1}{\cos {{40}^{\circ }}}\] ---(3).
Let us substitute equation (3) in equation (2).
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{1}{\cos {{40}^{\circ }}}-\tan {{40}^{\circ }} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{1}{\cos {{40}^{\circ }}}-\dfrac{\sin {{40}^{\circ }}}{\cos {{40}^{\circ }}} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{1-\sin {{40}^{\circ }}}{\cos {{40}^{\circ }}} \right) $ .
We know that $ {{\sin }^{2}}A+{{\cos }^{2}}A=1 $ , $ \sin 2A=2\sin A\cos A $ and $ \cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{{{\sin }^{2}}{{20}^{\circ }}+{{\cos }^{2}}{{20}^{\circ }}-2\sin {{20}^{\circ }}\cos {{20}^{\circ }}}{{{\cos }^{2}}{{20}^{\circ }}-{{\sin }^{2}}{{20}^{\circ }}} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{{{\left( \cos {{20}^{\circ }}-\sin {{20}^{\circ }} \right)}^{2}}}{\left( \cos {{20}^{\circ }}-\sin {{20}^{\circ }} \right)\left( \cos {{20}^{\circ }}+\sin {{20}^{\circ }} \right)} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{\cos {{20}^{\circ }}-\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}+\sin {{20}^{\circ }}} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{\cos {{20}^{\circ }}\left( 1-\dfrac{\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}} \right)}{\cos {{20}^{\circ }}\left( 1+\dfrac{\sin {{20}^{\circ }}}{\cos {{20}^{\circ }}} \right)} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{1-\tan {{20}^{\circ }}}{1+\tan {{20}^{\circ }}} \right) $ .
We know that $ \tan {{45}^{\circ }}=1 $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{\tan {{45}^{\circ }}-\tan {{20}^{\circ }}}{1+\tan {{45}^{\circ }}\tan {{20}^{\circ }}} \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \tan \left( {{45}^{\circ }}-{{20}^{\circ }} \right) \right) $ .
 $ \Rightarrow \alpha ={{\tan }^{-1}}\left( \tan {{25}^{\circ }} \right) $ .
 $ \Rightarrow \alpha ={{25}^{\circ }} $ .