Question

If ${{\tan }^{-1}}2$ , ${{\tan }^{-1}}3$ are two angles of a triangle, then the third angle is:
(A) ${{30}^{\circ }}$
(B) ${{45}^{\circ }}$
(C) ${{60}^{\circ }}$
(D) ${{75}^{\circ }}$

Answer 1

Hint: To solve this type of question we need to have the concept of the trigonometric function. To solve this question we are supposed to apply that the total sum of the angles of a triangle is equal to ${{180}^{\circ }}$. Then substituting the values in the formula we find the third angle of the triangle. The domain varies from $\left[ -\infty ,\infty \right]$.

Complete step by step solution:
The problem asks us to find the third angle of a triangle when two of the angles of the triangle are given as ${{\tan }^{-1}}2$ and ${{\tan }^{-1}}3$. We know that the total sum of the angles of a triangle is equal to ${{180}^{\circ }}$. Now considering the third angle of the triangle to be $\theta $ the sum of angle could be equated to find the third angle. The mathematical representation to write it would be:
$\Rightarrow ~{{\tan }^{-1}}2+{{\tan }^{-1}}3+\theta ={{180}^{\circ }}$
Taking $\theta $ to the left hand side of the equation, we get:
$\Rightarrow {{\tan }^{-1}}2+{{\tan }^{-1}}3={{180}^{\circ }}-\theta $
To solve the ${{\tan }^{-1}}2+{{\tan }^{-1}}3$we need to use the formula \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\] . Considering $x=2$ and $y=3$ substituting in the formula we get:
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{2+3}{1-2\times 3} \right)={{180}^{\circ }}-\theta \]
Calculating it further we get:
\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{-5} \right)={{180}^{\circ }}-\theta \]
\[\Rightarrow {{\tan }^{-1}}\left( -1 \right)={{180}^{\circ }}-\theta \]
Taking the trigonometric function $\tan $ both side we get:
\[\Rightarrow \tan \left( {{\tan }^{-1}}\left( \dfrac{5}{-5} \right) \right)=\tan \left( {{180}^{\circ }}-\theta \right)\]
Since ${{180}^{\circ }}-\theta $ angle is in the second quadrant. Trigonometric function$\tan $is negative in the second quadrant. So \[\tan \left( {{180}^{\circ }}-\theta \right)\] changes to $-\tan \theta $. Also $\tan \left( {{\tan }^{-1}}x \right)=x$ . Applying these two formula in the above equation we get:
$\Rightarrow -1=-\tan \theta $
Multiplying with $-1$ both side we get:
$\Rightarrow 1=\tan \theta $
Now, we know that $\tan \theta $ evaluates to $1$ when angle $\theta $ is $\dfrac{\pi }{4}$ , which in degree is ${{45}^{\circ }}$.
$\therefore $ If ${{\tan }^{-1}}2$ , ${{\tan }^{-1}}3$ are two angles of a triangle, then the third angle is $\left( B \right){{45}^{\circ }}$

Note: We need to remember the properties of trigonometric functions to solve this question. The domain varies from $\left[ -\infty ,\infty \right]$ , while the range of the inverse trigonometric function, ${{\tan }^{-1}}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. It should be known to us that the sum of angles in a triangle is ${{180}^{\circ }}$. In the second quadrant the trigonometric function $\tan$ is negative.