The correct option is A 2tanc2 tanα nbsp;+tanβ =sin ( α nbsp;+β nbsp; )/cosα nbsp;cosβ=2sin c/cos ( α nbsp;+β nbsp; )+cos ( α nbsp; β ...

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Conditional Identities

tanα +tanβ

Question

$tan α + tan β$

2 tan \frac{c}{2}

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tan c

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2 tan c

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2 tan \frac{c}{3}

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Solution

The correct option is A $2 tan \frac{c}{2}$
$tan α + tan β = \frac{sin (α + β)}{cos α cos β} = \frac{2 sin c}{cos (α + β) + cos (α - β)} = \frac{sin c}{{cos}^{2} \frac{c}{2} - {sin}^{2} (\frac{α - β}{2})}$
This is minimum when denominator is maximum i.e. when
${sin}^{2} (\frac{α - β}{2})$ is zero
$\Rightarrow \frac{2 sin \frac{c}{2} cos \frac{c}{2}}{{cos}^{2} \frac{c}{2}} = 2 tan \frac{c}{2}$

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Similar questions

tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1

If A+B+C=180°, prove that

tan(A/2)tan(B/2)+ tan(B/2)tan(C/2)+ tan(C/2)tan(A/2)= 1

A + B + C = π

tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1

x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z = tan (\frac{A - B}{2}) tan \frac{C}{2}

x + y + z = ?

A B C

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{C}{2} + tan \frac{A}{2} & tan \frac{A}{2} + tan \frac{B}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0

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