Question

If in a triangle $\vartriangle PQR$ , if $a\cdot \tan A+b\cdot \tan B=(a+b)\tan \left( \dfrac{A+B}{2} \right)$ , then prove that A = B.

Answer 1

Hint: This question is based on many concepts, so we will do this questions by using the various trigonometric formula, identities, and sine rule of the triangle, then by reframing the given equation in questions, we will prove that $\angle A = \angle B$

Complete step-by-step solution:

We have to prove that side A = Side B or we can say that we have to prove that $\vartriangle PQR$ is an isosceles triangle, as in a triangle when two angles or two sides are equal, then the triangle is isosceles.
Now, now in question it is given that $a\cdot \tan A+b\cdot \tan B=(a+b)\tan \left( \dfrac{A+B}{2} \right)$
So, opening brackets on right hand side, we get
$a\cdot \tan A+b\cdot \tan B=a\tan \left( \dfrac{A+B}{2} \right)+b\tan \left( \dfrac{A+B}{2} \right)$
Taking $a\cdot \tan \left( \dfrac{A+B}{2} \right)$ from right hand side to left hand side and $b\operatorname{tanB}$from left hand side to right hand side, we get
$a\cdot \tan A-a\tan \left( \dfrac{A+B}{2} \right)=b\tan \left( \dfrac{A+B}{2} \right)-b\cdot \tan B$
On, simplifying, we get
$a\cdot \left( \tan A-\tan \left( \dfrac{A+B}{2} \right) \right)=b\cdot \left( \tan \left( \dfrac{A+B}{2} \right)-\tan B \right)$
We know that, $\tan A=\dfrac{\sin A}{\cos A}$ , so
\[a\cdot \left( \dfrac{\sin A}{\cos A}-\dfrac{\sin \left( \dfrac{A+B}{2} \right)}{\cos \left( \dfrac{A+B}{2} \right)} \right)=b\cdot \left( \dfrac{\sin \left( \dfrac{A+B}{2} \right)}{\cos \left( \dfrac{A+B}{2} \right)}-\dfrac{\sin B}{\operatorname{cosB}} \right)\]
On taking L.C.M on both sides, we get
\[a\cdot \left( \dfrac{\sin A\cdot \cos \left( \dfrac{A+B}{2} \right)-\sin \left( \dfrac{A+B}{2} \right)\cdot \cos A}{\cos A\cdot \cos \left( \dfrac{A+B}{2} \right)} \right)=b\cdot \left( \dfrac{\sin \left( \dfrac{A+B}{2} \right)\cdot \cos B-\sin B\cdot \cos \left( \dfrac{A+B}{2} \right)}{\cos \left( \dfrac{A+B}{2} \right)\cdot \operatorname{cosB}} \right)\]
We know that, sin ( A - B ) = sinAcosB - sinBcosA
So, \[a\cdot \left( \dfrac{\sin \left( A-\dfrac{A+B}{2} \right)}{\cos A\cdot \cos \left( \dfrac{A+B}{2} \right)} \right)=b\cdot \left( \dfrac{\sin \left( \dfrac{A+B}{2}-B \right)}{\cos \left( \dfrac{A+B}{2} \right)\cdot \operatorname{cosB}} \right)\]
We also know that, sine rule in $\vartriangle ABC$says $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$
So, we can write a = ksinA and b = ksinB
\[ksinA\cdot \left( \dfrac{\sin \left( A-\dfrac{A+B}{2} \right)}{\cos A\cdot \cos \left( \dfrac{A+B}{2} \right)} \right)=ksinB\cdot \left( \dfrac{\sin \left( \dfrac{A+B}{2}-B \right)}{\cos \left( \dfrac{A+B}{2} \right)\cdot \operatorname{cosB}} \right)\]
On simplifying, we get
\[sinA\cdot \left( \dfrac{\sin \left( \dfrac{A-B}{2} \right)}{\cos A} \right)=sinB\cdot \left( \dfrac{\sin \left( \dfrac{A-B}{2} \right)}{\operatorname{cosB}} \right)\]
Again on simplifying,
\[\left( \dfrac{\operatorname{sinA}}{\cos A} \right)=\left( \dfrac{\operatorname{sinB}}{\operatorname{cosB}} \right)\]
Or, tanA = tanB
which implies, $\angle A = \angle B$, that is $\vartriangle PQR$ is isosceles triangle.

Note: As this question is lengthy try not to make any mistake and do not skip any step as it may confuse in the next steps which may make question confusing. One must know sine law of triangle which is in $\vartriangle ABC$, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$, trigonometric identities such as $\tan A=\dfrac{\sin A}{\cos A}$and $\tan A=\dfrac{1}{\cot A}$ while solving these proving based questions. The calculation must be done carefully.