If 2 cos- - x - 1x and 2 cos- - y - 1-y etc- then prove that - xy - yx - 2 cos - - -

xy = e^iθe^iθ = e^i(θ ϕ) ∴ y/x = (x/y)^ 1 = e^ i(θ ϕ) xy + yx = 2cos(θ ϕ). ...

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Sign of Trigonometric Ratios in Different Quadrants

If 2 cosθ ...

Question

If $2 c o s θ = x = \frac{1}{x} a n d 2 c o s ϕ = y + \frac{1}{y}$ etc., then prove that :
$\frac{x}{y} + \frac{y}{x} = 2 c o s (θ - ϕ)$

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2 c o s θ = x = \frac{1}{x} a n d 2 c o s ϕ = y + \frac{1}{y}

x y z . . . . + \frac{1}{x y z . . . .} = 2 c o s (θ + ϕ + . . . .)

2 cos θ = x + \frac{1}{x}

2 cos ϕ = y + \frac{1}{y}

x^{m} y^{n} + \frac{1}{x^{m} y^{n}} = 2 cos (m θ + n ϕ)

2 cos θ = x + \frac{1}{x}

2 cos ϕ = y + \frac{1}{y}

x = 2 \cos θ - \cos 2 θ and y = 2 \sin θ - \sin 2 θ

\frac{d y}{d x} = \tan (\frac{3 θ}{2})

$\frac{x}{a} c o s θ$ + $\frac{y}{b} s i n θ$ = 1

$\frac{x}{a} s i n θ$ - $\frac{y}{b} c o s θ$ = 1

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