# Conditional Identities (Tanθ) – Problems

## Conditional Identities (Tanθ) – Problems

1. If A + B + C = nπ, n Є Z that TanA + TanB + TanC = TanA TanB TanC.

Proof: We have

$$\operatorname{Tan}\left( A+B+C \right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}$$,

A + B + C = nπ

Tan (A + B + C) = tan(nπ)

= tan(π) = 0

$$0=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan A\tan C}$$,

TanA + TanB + TanC – TanA TanB TanC = 0

TanA + TanB + TanC = TanA TanB TanC

Hence proved.

Note: in ΔABC, A + B + C = π then TanA + TanB + TanC = TanA TanB TanC

2. If A + B + C = π, then Tan(A/2) tan(B/2) + tan(C/2) tan(B/2) + tan(C/2) tan(A/2) = 1

Proof: A + B + C = π

(A/2) + (B/2) + (C/2) = π/2

(A/2) + (B/2) = π/2 – (C/2)

$$\tan \left( \frac{A}{2}+\frac{B}{2} \right)=\tan \left( \frac{\pi }{2}-\frac{C}{2} \right)$$ $$\left( \because \ \tan \left( \frac{\pi }{2}-\frac{C}{2} \right)=\cot \left( \frac{C}{2} \right) \right)$$,

= cot(C/2)

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

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

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

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

$$\tan \left( \frac{A}{2} \right)\tan \left( \frac{C}{2} \right)+\tan \left( \frac{B}{2} \right)\tan \left( \frac{C}{2} \right)+\tan \left( \frac{A}{2} \right)\tan \left( \frac{B}{2} \right)=1$$,

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

Hence proved.