Question

tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1​

Answer 1

I think question is incomplete it is like that

In ΔABC prove that

tanA/2 tanB/2 + tanB/2 tanC/2 +

tanC/2 tanA/2 = 1

proof --->

In Δ ABC by angle sum property

A + B + C = 180°

Dividing both sides by 2

=> A/2 + B /2 + C/ 2 = 180° / 2

=> A / 2 + B / 2 = 90° - C / 2

Taking tan both sides we get

=> tan ( A/2 + B/2 )= tan (90° - C /2 )

We have some formulee

tanx + tany

tan (x + y ) = ---------------------

1 - tanx tany

tan (90° - θ ) = Cot θ

Applying it here

tanA/2 + tanB/2

=> ----------------------------- = Cot C/2

1 - tanA/2 tanB/2

tanA/2 + tanB/2 1

=> ------------------------------ = -------------

1 - tanA/2 tanB/2 tan C/2

=> tanC/2 (tanA/2 + tanB/2) = 1 - tanA/2

tanB/2

=> tanA/2 tanC/2 + tanB/2 tanC/2

= 1 - tanA/2 tanB/2

=> tanA/2 tanB/2 + tanB/2 tanC/2 +

tanC/2 tanA/2 = 1

Hence proved

Answer 2

Answer: tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1​

Step-by-step explanation:

In a triangle the sum of 3 angles is equal 180

Therefore, using this we can say that A+B+C=180

Just like this tanA means an angle so we can say that;

tanA+tanB+tanC=180

Therefore to get tanA/2 we divide the above equation by 2;

$tan(\frac{A}{2}+\frac{B}{2} +\frac{C}{2}) =tan\frac{180}{2}$

So we get the equation

$tan(\frac{A}{2}+\frac{B}{2} +\frac{C}{2}) =tan90$

From this equation we take tanC/2 to the right hand side then we get the equation;

$tan(\frac{A}{2}+\frac{B}{2} ) =tan(90 -\frac{C}{2})$

From the knowledge of circular functions we can say that tan(90-θ)=cotθ

Therefore,

$tan(\frac{A}{2}+\frac{B}{2} ) =cot\frac{C}{2}$

using AB formulae we can express tan(A/2+B/2) as;

$\frac{tan\frac{A}{2}+tan\frac{B}{2} }{1-tan\frac{A}{2}tan\frac{B}{2} }$

By this we get the equation;

$\frac{tan\frac{A}{2} +tan\frac{B}{2} }{1-tan\frac{A}{2} tan\frac{B}{2} }=cot\frac{C}{2}$

Now we have to remove 1-tanAtanB from the left hand side so, we multiply 1-tanA/2tanB/2 from both sides then 1-tanA/2tanB/2 gets cancelled from the left hand side and we get the equation;

$tan\frac{A}{2} +tan\frac{B}{2} =cot\frac{C}{2}({1-tan\frac{A}{2} tan\frac{B}{2} })$

cotC/2 can be written as $\frac{1}{tan\frac{C}{2} }$ then we can write this equation as ;

$tan\frac{A}{2} +tan\frac{B}{2} =\frac{1-tan\frac{A}{2}tan\frac{B}{2} }{tan\frac{C}{2} }$

Now we have to remove tanC/2 from the right hand side so we multiply both sides by tanC/2 and we get the equation;

${tan\frac{C}{2} }(tan\frac{A}{2} +tan\frac{B}{2} )=1-tan\frac{A}{2} tan\frac{B}{2}$

Now we remove the brackets by multiplying tanC/2 with tanA/2+tanB/2 and we get the equation;

$tan\frac{A}{2} {tan\frac{C}{2} }+tan\frac{B}{2}{tan\frac{C}{2} } =1-tan\frac{A}{2} tan\frac{B}{2}$

Now as the last step we take tanA/2tanB/2 to the left hand side and as we know when we take -tanA/2tanB/2 from one side to the other it becomes +tanA/2tanB/2

Now we get the equation;

$tan\frac{A}{2}tan\frac{B}{2} +tan\frac{B}{2} {tan\frac{C}{2} }+tan\frac{A}{2}{tan\frac{C}{2} } =1$

L.H.S=R.H.S