Question

If\[\dfrac{x}{\tan(\theta + \alpha)} = \dfrac{y}{tan(\theta + \beta)} = \dfrac{z}{\tan(\theta + \gamma )}\]then\[\sum \dfrac{x + y}{x - y} \sin^2 (\alpha - \beta ) ~ = \cdots\]

Answer 1

By the sum, it must mean this expression:\[\dfrac{x + y}{x-y}\sin^2 (\alpha - \beta) + \dfrac{y + z }{y - z} \sin^2 (\beta - \gamma) + \dfrac{x + z}{x - z} \sin^2(\alpha - \gamma)\]

Answer 2

Initial observations:\[\alpha - \beta = (\theta + \alpha) - (\theta + \beta)\]etc.

Answer 3

Let's do it like we solve ratio and proportion problems...\[\dfrac{x }{y} = \dfrac{\tan (\theta + \alpha)}{\tan(\theta + \beta)}\]Applying C and D\[\dfrac{x + y}{x - y} = \dfrac{\tan(\theta + \alpha) + \tan(\theta + \beta)}{\tan(\theta + \alpha) - \tan (\theta + \beta)}\]

Answer 4

\[\sin^2 (\alpha - \beta) = \dfrac{1 - \cos(2(\alpha - \beta))}{2}\]and\[\cos (2(\alpha - \beta)) = \dfrac{1 - \tan^2(\alpha - \beta)}{1 + \tan^2(\alpha + \beta)}\]Nope.

Answer 5

hartnn :(

Answer 6

@ganeshie8

Answer 7

Y'know what, scratch that. Assume \(\theta = 0 \), \(\alpha = 30^{\circ}\), \(\beta = 45^\circ\), \(\gamma = 60^{\circ}\) such that \(x = \frac{1}{\sqrt 3 }\), \(y = 1\), \(z = \sqrt 3\). =_=

Answer 8

\[\tan(\alpha - \beta) \]\[= \tan((\theta + \alpha) - (\theta + \beta) )\]\[ = \dfrac{\tan (\theta + \alpha) - \tan (\theta + \beta)}{1 + \tan(\theta + \alpha) \tan (\theta + \beta)}\]

Answer 9

Apply C and D change into sin and cos and use 2sinAsinB= cos (A-B) - cos(A+B) every term gets cancelled...zero!

Answer 10

Oh, so what I was missing was conversion into sin and cos. Thanks!

Answer 11

welcome :)