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Question: If \({{\tan }^{-1}}2\) , \({{\tan }^{-1}}3\) are two angles of a triangle, then the third angle is: ...

If tan⁡−12{{\tan }^{-1}}2 , tan⁡−13{{\tan }^{-1}}3 are two angles of a triangle, then the third angle is:
(A) 30∘{{30}^{\circ }}
(B) 45∘{{45}^{\circ }}
(C) 60∘{{60}^{\circ }}
(D) 75∘{{75}^{\circ }}

Explanation

Solution

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∘{{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⁡−12{{\tan }^{-1}}2 and tan⁡−13{{\tan }^{-1}}3. We know that the total sum of the angles of a triangle is equal to 180∘{{180}^{\circ }}. Now considering the third angle of the triangle to be θ\theta the sum of angle could be equated to find the third angle. The mathematical representation to write it would be:
⇒ tan⁡−12+tan⁡−13+θ=180∘\Rightarrow ~{{\tan }^{-1}}2+{{\tan }^{-1}}3+\theta ={{180}^{\circ }}
Taking θ\theta to the left hand side of the equation, we get:
⇒tan⁡−12+tan⁡−13=180∘−θ\Rightarrow {{\tan }^{-1}}2+{{\tan }^{-1}}3={{180}^{\circ }}-\theta
To solve the tan⁡−12+tan⁡−13{{\tan }^{-1}}2+{{\tan }^{-1}}3we need to use the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) . Considering x=2x=2 and y=3y=3 substituting in the formula we get:
⇒tan⁡−1(2+31−2×3)=180∘−θ\Rightarrow {{\tan }^{-1}}\left( \dfrac{2+3}{1-2\times 3} \right)={{180}^{\circ }}-\theta
Calculating it further we get:
⇒tan⁡−1(5−5)=180∘−θ\Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{-5} \right)={{180}^{\circ }}-\theta
⇒tan⁡−1(−1)=180∘−θ\Rightarrow {{\tan }^{-1}}\left( -1 \right)={{180}^{\circ }}-\theta
Taking the trigonometric function tan⁡\tan both side we get:
⇒tan⁡(tan⁡−1(5−5))=tan⁡(180∘−θ)\Rightarrow \tan \left( {{\tan }^{-1}}\left( \dfrac{5}{-5} \right) \right)=\tan \left( {{180}^{\circ }}-\theta \right)
Since 180∘−θ{{180}^{\circ }}-\theta angle is in the second quadrant. Trigonometric functiontan⁡\tan is negative in the second quadrant. So tan⁡(180∘−θ)\tan \left( {{180}^{\circ }}-\theta \right) changes to −tan⁡θ-\tan \theta . Also tan⁡(tan⁡−1x)=x\tan \left( {{\tan }^{-1}}x \right)=x . Applying these two formula in the above equation we get:
⇒−1=−tan⁡θ\Rightarrow -1=-\tan \theta
Multiplying with −1-1 both side we get:
⇒1=tan⁡θ\Rightarrow 1=\tan \theta
Now, we know that tan⁡θ\tan \theta evaluates to 11 when angle θ\theta is π4\dfrac{\pi }{4} , which in degree is 45∘{{45}^{\circ }}.
∴\therefore If tan⁡−12{{\tan }^{-1}}2 , tan⁡−13{{\tan }^{-1}}3 are two angles of a triangle, then the third angle is (B)45∘\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⁡−1x{{\tan }^{-1}}x is [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]. It should be known to us that the sum of angles in a triangle is 180∘{{180}^{\circ }}. In the second quadrant the trigonometric function tan⁡\tan is negative.