If A+B+C=180°, prove that

tan(A/2)tan(B/2)+ tan(B/2)tan(C/2)+ tan(C/2)tan(A/2)= 1

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Solution

A+B+C = 180
so A/2 + B/2 + C/2 = 180/2

or
A/2 + B/2 = (180/2- C/2)
take tan of both sides

tan (A/2 + B/2) = tan (180/2-C/2) = cot C/2 = 1/ tan C/2

or
(tan A/2 + tan B/2)/(1- Tan A/2 tan B/2) = 1/tan C/2

or
tan C/2 tan A/2 + tan C/2 tan B/2 = 1 - tan A/2 tan B/2
or
tan A/2 tan B/2 + tan B/2 tan C/2 + tan A/2 tan C/2 = 1
proved

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Similar questions

A + B + C = π

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

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

x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z = tan (\frac{A - B}{2}) tan \frac{C}{2}

x + y + z = ?

I n △ A B C, i f x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z tan (\frac{A - B}{2}) tan \frac{C}{2}, t h e n x + y + z

x = t a n \frac{B - C}{2} t a n \frac{A}{2}

y = t a n \frac{C - A}{2} t a n \frac{B}{2}

z = t a n \frac{A - B}{2} t a n \frac{C}{2}

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