Question

The value of $\tan 20{}^\circ +2\tan 50{}^\circ -\tan 70{}^\circ $ is:
a). 1
b). 0
c). $\tan 50{}^\circ $
d). none of these

Answer 1

Hint: To solve the above question we will first replace all the $\tan \theta $ by $\dfrac{\sin \theta }{\cos \theta }$ and then take the LCM by pairing two term separately and then by using $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$ and $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$ we will simplify it and at last we will use property $\sin \left( 90{}^\circ -\theta \right)=\cos \theta $, where $\theta $ is less than $90{}^\circ $, to further simplify it and get the required answer

Complete step by step answer:
Since, we do not know the value of any of them (i.e.$\tan 20{}^\circ $, $\tan 50{}^\circ $ and $\tan 70{}^\circ $) directly from the standard trigonometry table. So, we have to simplify $\tan 20{}^\circ +2\tan 50{}^\circ -\tan 70{}^\circ $ to find its value by using trigonometric properties.
So, we can write $\tan 20{}^\circ +2\tan 50{}^\circ -\tan 70{}^\circ $ as:
$=\tan 20{}^\circ +\tan 50{}^\circ +\tan 50{}^\circ -\tan 70{}^\circ $
Since, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ , so we will replace each $\tan \theta $ by $\dfrac{\sin \theta }{\cos \theta }$, then we will get:
$=\dfrac{\sin 20{}^\circ }{\cos 20{}^\circ }+\dfrac{\sin 50{}^\circ }{\cos 50{}^\circ }+\dfrac{\sin 50{}^\circ }{\cos 50{}^\circ }-\dfrac{\sin 70{}^\circ }{\cos 70{}^\circ }$
Now, we will take LCM of the denominator, then we will get:
$=\dfrac{\sin 20{}^\circ \cos 50{}^\circ +\sin 50{}^\circ \cos 20{}^\circ }{\cos 20{}^\circ \cos 50{}^\circ }+\dfrac{\sin 50{}^\circ \cos 70{}^\circ -\sin 70{}^\circ \cos 50{}^\circ }{\cos 50{}^\circ \cos 70{}^\circ }$
Now, we know that $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$ and similarly we also know that $\sin \left( A-B \right)=\sin A\cos B-\sin B\cos A$.
So, we can say that $\sin 20{}^\circ \cos 50{}^\circ +\sin 50{}^\circ \cos 20{}^\circ =\sin \left( 50{}^\circ +20{}^\circ \right)=\sin \left( 70{}^\circ \right)-----(1)$
Similarly, we can say that $\sin 50{}^\circ \cos 70{}^\circ -\sin 70{}^\circ \cos 50{}^\circ =\sin \left( 50{}^\circ -70{}^\circ \right)=\sin \left( -20{}^\circ \right)$.
Since, we know that $\sin \left( -\theta \right)=-\sin \theta $. So, we can write $\sin 50{}^\circ \cos 70{}^\circ -\sin 70{}^\circ \cos 50{}^\circ =\sin \left( -20{}^\circ \right)=-\sin 20{}^\circ ---(2)$
Now, after putting (1) and (2) value in $\dfrac{\sin 20{}^\circ \cos 50{}^\circ +\sin 50{}^\circ \cos 20{}^\circ }{\cos 20{}^\circ \cos 50{}^\circ }+\dfrac{\sin 50{}^\circ \cos 70{}^\circ -\sin 70{}^\circ \cos 50{}^\circ }{\cos 50{}^\circ \cos 70{}^\circ }$, we will get:
$=\dfrac{\sin 70{}^\circ }{\cos 20{}^\circ \cos 50{}^\circ }-\dfrac{\sin 20{}^\circ }{\cos 50{}^\circ \cos 70{}^\circ }------(3)$
Now, we know that $\sin \left( 90{}^\circ -\theta \right)=\cos \theta $ where $\theta $ is less than $90{}^\circ $, so we can write $\cos 20{}^\circ $ as $\sin \left( 90{}^\circ -20{}^\circ \right)$. Similarly, we can also write $\cos 70{}^\circ $ as $\sin \left( 90{}^\circ -70{}^\circ \right)$ .
Now, after putting $\sin \left( 90{}^\circ -20{}^\circ \right)$ in place of $\cos 20{}^\circ $ and $\sin \left( 90{}^\circ -70{}^\circ \right)$ in place of $\cos 70{}^\circ $in (3), we will get:
$=\dfrac{\sin 70{}^\circ }{\sin \left( 90{}^\circ -20{}^\circ \right)\cos 50{}^\circ }-\dfrac{\sin 20{}^\circ }{\cos 50{}^\circ \sin \left( 90{}^\circ -70{}^\circ \right)}$
$=\dfrac{\sin 70{}^\circ }{\sin 70{}^\circ \cos 50{}^\circ }-\dfrac{\sin 20{}^\circ }{\cos 50{}^\circ \sin 20{}^\circ }$
$=\dfrac{1}{\cos 50{}^\circ }-\dfrac{1}{\cos 50{}^\circ }$
= 0
This is our required solution.

So, the correct answer is “Option b”.

Note: We can also solve the above question alternatively by splitting $\tan 70{}^\circ $ as $\tan \left( 50{}^\circ +20{}^\circ \right)=\dfrac{\tan 50{}^\circ +\tan 20{}^\circ }{1-\tan 50{}^\circ \tan 20{}^\circ }$ so by putting $\tan 50{}^\circ +\tan 20{}^\circ =\tan 70{}^\circ \left( 1-\tan 50{}^\circ \tan 20{}^\circ \right)$ in $\tan 20{}^\circ +2\tan 50{}^\circ -\tan 70{}^\circ $.
So, we will get: $\tan 20{}^\circ +2\tan 50{}^\circ -\tan 70{}^\circ $= \[\tan 50{}^\circ +\tan 70{}^\circ \left( 1-\tan 50{}^\circ \tan 20{}^\circ \right)-\tan 70{}^\circ \]
$=\tan 50{}^\circ +\tan 70{}^\circ -\tan 70{}^\circ \tan 50{}^\circ \tan 20{}^\circ -\tan 70{}^\circ $
Now, we will use the property $\cot \theta =\tan \left( 90{}^\circ -\theta \right)$ and write $\tan 70{}^\circ =\cot 20{}^\circ $
Hence, we will get: $=\tan 50{}^\circ -\cot 20{}^\circ \tan 50{}^\circ \tan 20{}^\circ $
$\Rightarrow \tan 50{}^\circ -\tan 50{}^\circ =0$
Students are required to memorize all the trigonometric properties and use them correctly and in such a way that the form given in question gets simplified into a simpler form and the form given in the option.