Question

tan 15° =????????????????? ​

Answer 1

HEYA MATE !

Tan 15° = Tan (45° - 30°)

__tan(A-B)=(tanA-tanB)/(1+tanA.tanB)__

Tan 15° = (2)-(√3)

hope it helps

Answer 2

$\small \bf \blue{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \purple{ \huge \: solution} \\ \\ \small \bf \pink{tan \: 15°} \\ \\ \small \bf \red{as \: we \: can \: write} \\ \\ \small \bf \pink{tan15 °= tan(60° - 45°)} \\ \\ \small \bf \pink{taking \: \: tan(60° - 45°)} \\ \\ \small \bf \orange{using \: this \: identity} \\ \\ \small \bf \pink{tan( \alpha - \beta ) = \frac{tan \alpha - tan \beta }{1 + tan \alpha \times tan \beta } } \\ \\ \small \bf \pink{tan(60° -45°) = \frac{tan60° - tan45°}{1 + tan60 °\times tan45°} } \\ \\ \small \bf \pink{tan15° = \frac{ \sqrt{3} - 1}{1 + \sqrt{3 } \times 1} } \\ \\ \small \bf \pink{tan15 °= \frac{ \sqrt{3} - 1 }{1 + \sqrt{3} } } \\ \\ \small \bf \pink{ tan15 °= \frac{( \sqrt{3} - 1)( \sqrt{3} - 1) }{( \sqrt{3} + 1)( \sqrt{3} - 1) } } \\ \\ \small \bf \pink{ tan15 °= \frac{( \sqrt{3} - 1)^{2} }{( \sqrt{3})^{2} - {(1)}^{2} } } \\ \\ \small \bf \pink{tan15° = \frac{ \sqrt{3 }^{2} - 2 \times \sqrt{3} \times 1 + {1}^{2} }{3 - 1} } \\ \\ \small \bf \pink{tan15 °= \frac{3 - 2 \sqrt{3} + 1 }{2} } \\ \\ \small \bf \pink{tan15° = \frac{3 + 1 - 2 \sqrt{3} }{2} } \\ \\ \small \bf \pink{tan15 °= \frac{4 - 2 \sqrt{3} }{2} } \\ \\ \small \bf \pink{tan15° = \frac{2(2 - \sqrt{3} )}{2} } \\ \\ \small \bf \pink{tan15° = 2 - \sqrt{3} \: \: ans }$