Question

If 6x = 90, then tan3x-tan 2x-tanx /tan3x.
tan2x.tanx​

Answer 1

$\huge\mathfrak{\red{\fbox{\fbox{\underline{\underline{ANSWER}}}}}}$

$6x = 90 \\ x = \dfrac{90}{6} \\ x = 15$

$\dfrac{ \tan(45) - \tan(30) - \tan(15) }{ \tan(45) \times \tan(30) \times \tan(15)} \\ \\ \dfrac{1 - \dfrac{1}{ \sqrt{3}} - \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }}{1 \times \dfrac{1}{ \sqrt{3} } \times \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } } \\ \\ \dfrac{ \dfrac{ \sqrt{3}( \sqrt{3} + 1) - ( \sqrt{3} + 1) - \sqrt{3}( \sqrt{3} - 1) }{ \sqrt{3}( \sqrt{3} + 1)} }{ \dfrac{ \sqrt{3} - 1 }{ \sqrt{3}( \sqrt{3} + 1) } }$

$\dfrac{3 + \sqrt{3} - \sqrt{3} - 1 - 3 + 1 }{ \sqrt{3} - 1 } \\ \\ \dfrac{0}{ \sqrt{3} - 1} \\ \\ 0$

Hope it helps