Question

Solve this
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Answer 1

        $\text{We must find }tan15$

        $\text{We can write }tan15 \text{ as }tan(45-30)$

        $\text{We know that }tan(A-B) = \dfrac{tanA-tanB}{1 + tanAtanB}$

$\implies \: tan(45 - 30) = \dfrac{tan45 - tan30}{1 + tan45.tan30}$

$\implies \: tan15 = \dfrac{1 - \dfrac{1}{\sqrt3}}{1 + 1.\dfrac{1}{\sqrt3}}$

$\implies \: tan15 = \dfrac{\dfrac{\sqrt3-1}{\sqrt3}}{\dfrac{\sqrt3 + 1}{\sqrt3}}$

$\implies \: \boxed{tan15 = \dfrac{\sqrt3 - 1}{\sqrt3 + 1}}$