Question

Verify the formula (tan(α - β) = $\frac{tan \alpha - tan \beta }{1 + tan \alpha tan \beta}$) for: α = $\frac{\pi }{3}$ and β = $\frac{\pi }{6}$

Answer 1

$\Large{\underline{\underline{\mathfrak{\green{\bf{Question}}}}}}$

Verify the formula tan(α - β) = $\sf{\:\dfrac{\tan \alpha - \tan \beta }{1 + \tan \alpha \tan \beta}}$ for:- α = $\sf{\:\dfrac{\pi }{3}}$ and β = $\sf{\:\dfrac{\pi }{6}}$

$\Large{\underline{\underline{\mathfrak{\green{\bf{Prove}}}}}}$

$:\mapsto\sf{\:\tan(\alpha-\beta)\:=\:\dfrac{\tan \alpha - \tan \beta}{1+\tan \alpha\times\tan \beta}}$

$\:\:\:\:\:\small\pink{\sf{\:( keep\:value\:of\:\alpha\:and\beta)}}$

$:\mapsto\sf{\:\tan(\dfrac{\pi}{3}-\dfrac{\pi}{6})\:=\:\dfrac{\tan \dfrac{\pi}{3}-\tan \dfrac{\pi}{6}}{1+\tan \dfrac{\pi}{3}\times\tan \dfrac{\pi}{6}}}$

$\:\:\:\:\small\pink{\sf{\:( \tan \dfrac{\pi}{3}\:=\:\sqrt{3} )}}$

$\:\:\:\:\small\pink{\sf{\:( \tan \dfrac{\pi}{6}\:=\:\dfrac{1}{\sqrt{3}} )}}$

$:\mapsto\sf{\:\tan(\dfrac{2 \pi-\pi}{6})\:=\:\dfrac{\sqrt{3}-\dfrac{1}{\sqrt{3}}}{1+\sqrt{3}\times\dfrac{1}{\sqrt{3}}}}$

$:\mapsto\sf{\:\tan(\dfrac{\pi}{6})\:=\:\dfrac{3-1}{\sqrt{3}(1+1)}}$

$:\mapsto\sf{\:(\dfrac{1}{\sqrt{3}})\:=\:\dfrac{\cancel{2}}{\cancel{2}\sqrt{3}}}$

$:\mapsto\sf{\:\dfrac{1}{\sqrt{3}}\:=\:\dfrac{1}{\sqrt{3}}}$

$\Large\sf{\bf{\:L.H.S.\:=\:R.H.S.}}$

That't proved

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Answer 2

Answer:

$\large\boxed{\sf{\dfrac{1}{\sqrt{3}}}}$

Step-by-step explanation:

Given formula to verify :

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

Also, given that,

$\alpha = \dfrac{\pi}{3} \: \: \: \: \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \: \: \: \: \: \beta = \dfrac{\pi}{6}$

For verification:

L.H.S

$= \tan( \alpha - \beta ) \\ \\ = \tan( \frac{\pi}{3} - \dfrac{\pi}{6} ) \\ \\ = \tan( \dfrac{\pi}{3} ) \\ \\ = \dfrac{1}{ \sqrt{3} }$

R.H.S

$= \dfrac{ \tan\alpha - \tan \beta }{1 + \tan \alpha \tan \beta } \\ \\ = \dfrac{ \tan \frac{\pi}{3} - \tan \frac{\pi}{6} }{1 + \tan( \frac{\pi}{3} ) \tan( \frac{\pi}{6} ) } \\ \\ = \dfrac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + ( \sqrt{3} \times \frac{1}{ \sqrt{3} }) } \\ \\ = \dfrac{ \frac{3 - 1}{ \sqrt{3} } }{1 + 1} \\ \\ = \dfrac{ \frac{2}{ \sqrt{3} } }{2} \\ \\ = \dfrac{1}{\sqrt{3} }$

Thus, L.H.S = R.H.S

Hence, Verified.