Question

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$\huge \: hlo \: guys$


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$\huge \: gm$


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$find \: the \: value \: of \: \: tan \: 15°$

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Answer 1

$\huge \bold \red{answer - }$

Let tan(15°) = tan(45°-30°)

We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)

⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°) = {1- (1/√3)} / {1+(1/√3)}

∴ tan15° = (√3 - 1) / (√3 + 1)

Answer 2

$\small \bf \blue{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \blue{ \huge \: solution} \\ \\ \small \bf \blue{tan \: 15} \\ \\ \small \bf \blue{as \: we \: can \: write} \\ \\ \small \bf \blue{tan(60 - 45)} \\ \\ \small \bf \blue{using \: this \: formula} \\ \\ \small \bf \blue{tan(a - b) = \frac{tan \: a - tan \: b}{1 + tan \: a \: tan \: b} } \\ \\ \small \bf \blue{tan(60° - 45°) = \frac{tan \: 60° - tan \: 45°}{1 + tan \: 60° \times tan \: 45°} } \\ \\ \small \bf \blue{tan \: 15° = \frac{ \sqrt{3} - 1}{1 + \sqrt{3} - 1 } } \\ \\ \small \bf \blue{tan \: 15 °= \frac{ \sqrt{3} - 1}{ \sqrt{3} } } \\ \\ \small \bf \blue{tan \: 15° = \frac{( \sqrt{3} - 1) \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } } \\ \\ \small \bf \blue{tan \: 15° = \frac{3 - \sqrt{3} }{3} \: \: ans}$