If asin^2 θ + bcos^2 θ = c, bsin^2 φ + acos^2 φ = d and atanθ = btanφ then prove that (1/a) + (1/b) = (1/c) + (1/d) .


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Question Number 206500 by MATHEMATICSAM last updated on 16/Apr/24

$If a \sin^{2} θ + b \cos^{2} θ = c, b \sin^{2} ϕ + a \cos^{2} ϕ = d$ $and a \tan θ = b \tan ϕ then prove that$ $\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d} .$
	Answered by Berbere last updated on 17/Apr/24

	$s u p p s e c o s (θ) . c o s (ϕ) \neq 0$ $(1) \Leftrightarrow a \tan^{2} (θ) + b = c (1 + \tan^{2} (θ)) \Leftrightarrow {t a n}^{2} (θ) = \frac{c - b}{a - c}$ $(2) \Leftrightarrow \frac{d - a}{b - d} = \tan^{2} (ϕ)$ $\frac{b^{2}}{a^{2}} = \frac{(c - b) (b - d)}{(a - c) (d - a)} \Leftrightarrow (1 - \frac{c}{a}) (\frac{d}{a} - 1) = (\frac{c}{b} - 1) (1 - \frac{d}{b})$ $\Leftrightarrow \frac{c + d}{a} - \frac{c d}{a^{2}} = \frac{c + d}{b} - \frac{c d}{b^{2}}$ $\Leftrightarrow (c + d) (\frac{1}{a} - \frac{1}{b}) + c d (\frac{1}{b^{2}} - \frac{1}{a^{2}}) = 0$ $\Leftrightarrow (c + d) (\frac{1}{a} - \frac{1}{b}) - c d (\frac{1}{a} - \frac{1}{b}) (\frac{1}{a} + \frac{1}{b}) = 0$ $\Leftrightarrow (\frac{1}{a} - \frac{1}{b}) (c + d - c d (\frac{1}{a} + \frac{1}{b})) = 0$ $i f a = b \Rightarrow c = d = a = b \Rightarrow T r u e$ $i f c + d - c d (\frac{1}{a} + \frac{1}{b}) = 0 \Rightarrow c d (\frac{1}{a} + \frac{1}{b}) = c + d$ $You can't use 'macro parameter character #' in math mode$ $\Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{d} + \frac{1}{c}$ $c a s e (2) c o s (θ) c o s (ϕ) = 0 i m p o s s i b l d u e t o t h e c o n d i t i o n$ $(a \tan (θ) = b \tan (ϕ))$ $You can't use 'macro parameter character #' in math mode$ $\frac{1}{a} + \frac{1}{b} = \frac{1}{d} + \frac{1}{c}$
	Answered by Frix last updated on 17/Apr/24

	$One possible path :$ $\cos^{2} θ = \frac{a - c}{a - b} \land \cos^{2} ϕ = - \frac{b - d}{a - b}$ $a \tan θ = b \tan ϕ = x \Rightarrow$ $θ = \tan^{- 1} \frac{x}{a} \land ϕ = \tan^{- 1} \frac{x}{b}$ $\Rightarrow$ $x^{2} = - \frac{a^{2} (b - c)}{a - c} = - \frac{(a - d) b^{2}}{b - d}$ $\Rightarrow$ $d = - \frac{a b c}{a b - a c - b d}$ $\Rightarrow$ $\frac{1}{d} = \frac{1}{a} + \frac{1}{b} - \frac{1}{c}$ $\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d}$
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