If A-B-C- prove that tanA2tanB2-tanB2tanC2-tanC2tanA2-1-

It is given that A + B + C = π , then, A2 + B2 + C2 = π2 A2 + B2 = π2 C2 tan( A2 + B2) = tan( π2 C2) tanA2 + tanB21 tanA2tanB2 = C ...

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Use of Trigonometric Table

If A+B+C=π,...

Question

If $A + B + C = π$ , prove that $tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1$ .

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Solution

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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

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

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 = ?

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}

A B C

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{C}{2} + tan \frac{A}{2} & tan \frac{A}{2} + tan \frac{B}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0

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