Question

Find the value of$\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ$.
A) 0
B) $\dfrac{1}{\begin{array}{l}2\\\\\end{array}}$
C) -1
D) 1

Answer 1

Here we will use the concept of trigonometry. We will first use the formula of tangent of the sum of angles to form the given expression. Then we will substitute the angles of the given expression in the formula to simplify the formula and get the required answer

Formula used:
We will use the formula of the tangent of the sum of two angles is given by the formula as follows-$\tan (A + B)$=$\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

Complete step by step solution:
Now to find the value of $\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ$, we will take $A = 100^\circ$ and $B = 125^\circ$.
Thus by putting values for $A$ and $B$ in the above formula we will get,
$\tan (100^\circ + 125^\circ ) = \dfrac{{\tan 100^\circ + \tan 125^\circ }}{{1 - \tan 100^\circ \tan 125^\circ }}$
$\tan 225^\circ = \dfrac{{\tan 100^\circ + \tan 125^\circ }}{{1 - \tan 100^\circ \tan 125^\circ }}$
Now we know that, $\tan 225^\circ = 1$.
Thus by putting value of $\tan 225^\circ$ in the above equation, we get
$1 = \dfrac{{\tan 100^\circ + \tan 125^\circ }}{{1 - \tan 100^\circ \tan 125^\circ }}$
Multiplying both sides of the above equation by $\left( {1 - \tan 100^\circ \tan 125^\circ } \right)$ we will get,
$\left( {1 - \tan 100^\circ \tan 125^\circ } \right)1 = \dfrac{{\tan 100^\circ + \tan 125^\circ }}{{1 - \tan 100^\circ \tan 125^\circ }}\left( {1 - \tan 100^\circ \tan 125^\circ } \right)$
$1 - \tan 100^\circ \tan 125^\circ = \tan 100^\circ + \tan 125^\circ$
Now adding $\tan 100^\circ \tan 125^\circ$ on both sides of the equation, we get
$1 = \tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ$
Therefore we get the value of $\tan 100^\circ + \tan 125^\circ + \tan 100^\circ \tan 125^\circ$ is 1

Hence option (D) is the correct option.

Note:
Tangent function – In any right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side.
The tangent of an angle is defined to be its sine divided by its cosine: $\tan A = \dfrac{{\sin A}}{{\cos A}}$.
Formula $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ can be derived from $\dfrac{{\sin \left( {A + B} \right)}}{{\cos \left( {A + B} \right)}}$ where,
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Here we used $\tan 225^\circ = 1$ because we can write$\tan 225^\circ$ as $\tan \left( {\pi + 45^\circ } \right)$ which is equal to $\tan 45^\circ$.
And we know that $\tan 45^\circ = 1$.
Thus, we took $\tan 225^\circ$ is equal to 1.