First slide

Compound angles

Question

If $\tan^{2} \frac{π - A}{4} + \tan^{2} \frac{π - B}{4} + \tan^{2} \frac{π - C}{4} = 1,$ then $∆$ ABC is

Moderate

A

equilateral
B

isosceles
C

scalene
D

none of these

Solution

$α = \frac{π - A}{4}, β = \frac{π - B}{4}, γ = \frac{π - C}{4}$
$\begin{array}{l} \Rightarrow α + β + γ = \frac{π}{2} \\ \Rightarrow \sum \tan αtan β = 1 \\ i f x = \tan α, y = \tan β, z = \tan γ \\ \Rightarrow \sum \tan^{2} α = 1 = \sum \tan αtan β \\ x^{2} + y^{2} + z^{2} = x y + y z + z x \Rightarrow {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} = 0 \\ \Rightarrow \tan α = \tan β = \tan γ \end{array}$ f

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