Question

Using tan (A + B) =
(tan A+tan B)/(1-tan A*tan B, show that tan 75º = 2 + √3.​

Answer 1

$\huge{\mathrm{\underline{\underline\pink{ANSWER}}}}$

we know that

$\tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ). \tan( \beta ) }$

and,

$\tan(75°)$

can write as

$\tan(45° + 30°)$

we know that

$\tan(45°) = 1 \\ and \\ \tan(30°) = \frac{1}{ \sqrt{3} }$

${\mathrm{\underline{\underline\green{using \: \tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha) . \tan( \beta ) } }}}}$

now,

$\alpha = 45° \\ \beta = 30°$

substitute in the above given formula

$\tan(45° + 30°) = \frac{ \tan(45°) + \tan(30°) }{1 - \tan( 45°) . \tan(30°) }$

$\tan(75°) = \frac{1 + \frac{1}{ \sqrt{3} } }{1 - (1)( \frac{1}{ \sqrt{3} } )} \\ \tan(75°) = \frac{ \frac{ \sqrt{3} + 1 }{ \sqrt{3} } }{1 - \frac{1}{ \sqrt{3} } } \\ \\ \tan(75°) = \frac{ \frac{ \sqrt{3} + 1}{ \sqrt{3} } }{ \frac{ \sqrt{3} - 1}{ \sqrt{3} } } \\ \\ \tan(75°) = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1}$

rationalize the denominator

$\tan(75°) = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1} \\ \\ \tan(75°) = \frac{ {( \sqrt{3} + 1)}^{2} }{ {( \sqrt{3}) }^{2} - {(1)}^{2} } \\ \\ \tan(75°) = \frac{3 + 2 \sqrt{3} + 1 }{3 - 1} \\ \\ \tan(75°) = \frac{4 + 2 \sqrt{3} }{2} \\ \\ \tan(75°) = \frac{2(2 + \sqrt{3}) }{2}$

${\mathrm{\underline\green{ \tan(75°) = 2 + \sqrt{3} }}}$

HENCE PROVED

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