Question

prove that :

tan [pi/4 + 1/2 cos-1 a/b] + tan[pi/4 - 1/2cos-1 a/b] = 2b/a

Answer 1

☆☆ranshsangwan☆

tan( pi/4+1/2cos^-1 a/b)+tan(pi/4+1/2cos^-1a/b)

=2b/a..

Answer 2

${\huge{\underline{\underline{\mathbb{SOLUTION:-}}}}}$

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The equation to be proved true is this.

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$tan( \frac{\pi}{4} + \frac{1}{2}cos {}^{-1} \frac{a}{b} ) + tan( \frac{\pi}{4} - \frac{1}{4} cos {}^{-1} \frac{a}{b} ) = \frac{2b}{a}$

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To make the manipulation of the equations convenient, introduce one simplification substitution.

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$= \frac{1}{2} cos {}^{ - 1} \frac{a}{b} = θ</p><p>$

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Work on the left-hand side of the equation.

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$tan( \frac{\pi}{4} + θ ) + tan( \frac{\pi}{4} + θ ) = \frac{1\: +\:tan\:θ} {1\:-\:tan\:θ}+ \frac{1\: +\:tan\:θ} {1\:-\:tan\:θ}$

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$= tan( \frac{\pi}{4} + θ ) + tan( \frac{\pi}{4} - θ )$

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$=\frac{ (1 + tan\:θ) {}^{2} + (1 - tan\:θ) {}^{2}}{1 -tan {}^{2}θ }$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ)$

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$= \frac{1 + 2\:tan\:θ + {tan}^{2}θ + 1 - 2\:tan\:θ + tan {}^{2}θ}{1 - tan {}^{2}\:θ}$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ)$

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$= \frac{2(1 + {tan}^{2}θ) }{1-tan {}^{2}θ}$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ)$

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$= \frac{2 {sec}^{2}θ }{1-( {sec}^{2} θ - 1)}$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ)$

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$= \frac{2 {sec}^{2}θ }{2- {sec}^{2} θ}$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ)$

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$= \frac{2}{2- {cos}^{2} θ - 1}$

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$= tan( \frac{\pi}{4} + θ) + tan( \frac{\pi}{4} - θ = \frac{2}{cos2θ} )$

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Put θ back.

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$= tan( \frac{\pi}{4} + \frac{1}{2} {cos}^{-1} \frac{a}{b} ) + tan$

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$= (\frac{\pi}{4} - \frac{1}{2} {cos}^{-1} \frac{a}{b} ) = \frac{2}{cos2( \frac{1}{2} {cos}^{-1} \frac{a}{b}) }$

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$= tan( \frac{\pi}{4} + \frac{1}{2} {cos}^{-1} \frac{a}{b} ) + tan$

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$= (\frac{\pi}{4} - \frac{1}{2} {cos}^{-1} \frac{a}{b} ) = \frac{2}{ \frac{a}{b} }$

$= tan( \frac{\pi}{4} + \frac{1}{2} {cos}^{-1} \frac{a}{b} ) + tan$

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$= (\frac{\pi}{4} - \frac{1}{2} {cos}^{-1} \frac{a}{b} )$

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$=\frac{2b}{a}$

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That does it.