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Question: If\(A+B+C=\pi \), then using trigonometric identities, prove that \(\tan A+\tan B+\tan C=\tan A\ta...

IfA+B+C=πA+B+C=\pi , then using trigonometric identities, prove that
tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A+\tan B+\tan C=\tan A\tan B\tan C

Explanation

Solution

Hint: In this question, the sum of the angles is given and we have to find the relation between the tangents of these angles. Therefore, using the information, we can express the sum of two angles in terms of the other angle and π\pi and then use the formula for the tangent of the sum of angles to obtain the expression given in the question.

Complete step-by-step answer:
In the question, the sum of the three angles A, B and C is given to be
A+B+C=πA+B+C=\pi
Therefore,
A+B=π−C..............(1.1)A+B=\pi -C..............(1.1)
Taking the tangent on both sides we get,
tan⁡(A+B)=tan⁡(π−C)..............(1.2)\tan \left( A+B \right)=\tan \left( \pi -C \right)..............(1.2)
Now, we the formula for tangent of sum of two angles is given by
tan⁡(a+b)=tan⁡(a)+tan⁡(b)1−tan⁡(a)tan⁡(b).......(1.2)\tan \left( a+b \right)=\dfrac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)}.......(1.2)
And the tangent of two angles is given by
tan⁡(a−b)=tan⁡(a)−tan⁡(b)1+tan⁡(a)tan⁡(b).......(1.3)\tan \left( a-b \right)=\dfrac{\tan (a)-\tan (b)}{1+\tan (a)\tan (b)}.......(1.3)
Therefore, using equations (1.2) and (1.3) in equation (1.2), we obtain
tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)=tan⁡(π)−tan⁡(C)1+tan⁡(π)tan⁡(C).......(1.4)\dfrac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}=\dfrac{\tan (\pi )-\tan (C)}{1+\tan (\pi )\tan (C)}.......(1.4)

Now, as the angle π\pi is equal to 180∘{{180}^{\circ }}, therefore its tangent is given by
tan⁡(π)=sin⁡(π)cos⁡(π)=0−1=0\tan (\pi )=\dfrac{\sin (\pi )}{\cos (\pi )}=\dfrac{0}{-1}=0
Therefore using the value of tan⁡(π)\tan (\pi ) in equation (1.4), we obtain
tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)=0−tan⁡(C)1+0×tan⁡(C)=−tan⁡(C).......(1.5)\dfrac{\tan (A)+\tan (B)}{1-\tan (A)\tan (B)}=\dfrac{0-\tan (C)}{1+0\times \tan (C)}=-\tan (C).......(1.5)
Cross multiplying the numerator and denominator, we obtain
tan⁡(A)+tan⁡(B)=−tan⁡(C)(1−tan⁡(A)tan⁡(B))=−tan⁡(C)+tan⁡(A)tan⁡(B)tan⁡(C) ⇒tan⁡(A)+tan⁡(B)+tan⁡(C)=tan⁡(A)tan⁡(B)tan⁡(C) \begin{aligned} & \tan (A)+\tan (B)=-\tan (C)\left( 1-\tan (A)\tan (B) \right)=-\tan (C)+\tan (A)\tan (B)\tan (C) \\\ & \Rightarrow \tan (A)+\tan (B)+\tan (C)=\tan (A)\tan (B)\tan (C) \\\ \end{aligned}
Which is exactly the equation that we wanted to prove. Thus, we have successfully proved that
tan⁡(A)+tan⁡(B)+tan⁡(C)=tan⁡(A)tan⁡(B)tan⁡(C)\tan (A)+\tan (B)+\tan (C)=\tan (A)\tan (B)\tan (C)

Note: To prove the equation, we used the given relation that the sum of the angles is 180∘{{180}^{\circ }}. Now, the sum of the angles of a triangle is also 180∘{{180}^{\circ }}. Therefore, the same method will also be applicable to prove the relation for the angles of a triangle. Also, in equation (1.2), we could have also used the relation that
tan⁡(π−θ)=−tan⁡(θ)\tan (\pi -\theta )=-\tan \left( \theta \right) to simplify the RHS directly without using equation (1.3).