Question

The value of $\tan 203{}^\circ +\tan 22{}^\circ +\tan 203{}^\circ \cdot \tan 22{}^\circ $ is
(a) $-1$
(b) 0
(c) 1
(d) 2

Answer 1

Hint: First, we have to do addition of 2 angles given which will be $203{}^\circ +22{}^\circ =225{}^\circ $ . Then, we have to split 225 in such a way that $\tan \left( \pi +\theta \right)=\tan \theta $ . After this, we have to use the formula of addition of two angles given as $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . This, on solving, will give us the answer.

Complete step-by-step answer:
Here, we can see that in the only two angles are there i.e. $203{}^\circ $ and $22{}^\circ $ . Also, all the terms are in one of the trigonometric functions which is tan. So, first we will add both the angles and we will get $203{}^\circ +22{}^\circ =225{}^\circ $ .
So, here we can split 225 in such a way that something plus \[\theta \] i.e. \[\tan \theta \] should be known to us so that it is easy for simplification. It is written as $\left( 180+45 \right)=225$ . So, we know that tan after 180 degrees according to the graph comes in the third quadrant where all the values of tan functions are positive.
$\tan \left( \pi +\theta \right)=\tan \left( 180+45 \right)=\tan 45=1$ …………………………..(1)
Now, we will use the formula $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ where a is 203, b is 22. On substituting the values, we get
 $\tan \left( 203+22 \right)=\dfrac{\tan 203+\tan 22}{1-\tan 203\tan 22}$
Using value of equation (1), we get
$\tan \left( 225 \right)=1=\dfrac{\tan 203+\tan 22}{1-\tan 203\tan 22}$
On further simplification, we get
$1-\tan 203\tan 22=\tan 203+\tan 22$
Taking constant term on LHS and rest terms on RHS, we get
$1=\tan 203+\tan 22+\tan 203\tan 22$
Thus, $\tan 203{}^\circ +\tan 22{}^\circ +\tan 203{}^\circ \cdot \tan 22{}^\circ $ is 1.
Hence, option (c) is the correct answer.

Note: Another approach to this problem is by taking $\tan \left( 225 \right)=\tan \left( 270-45 \right)$ . This will also result in the same answer as done in the solution part. There will be no change in the equation $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . Therefore, calculating with this approach we get
$\Rightarrow \tan \left( 225 \right)=\tan \left( 270-45 \right)=\dfrac{\tan 203+\tan 22}{1-\tan 203\tan 22}$
$\Rightarrow \tan 45=1=\dfrac{\tan 203+\tan 22}{1-\tan 203\tan 22}$
On simplification, we get
$\Rightarrow 1=\tan 203+\tan 22+\tan 203\tan 22$ .
Thus, the answer will be the same.