Question

Prove that tan70°=tan20°+2tan50°

Answer 1

I have already answered this question before, So I have copy pasted it.

Given LHS = tan 70.

tan70=(tan50+tan20)/(1-tan50tan20)

tan70(1-tan50tan20)=tan50+tan20

tan70-tan70tan50tan20=tan50+tan20

tan70-tan(90-20)tan50tan20=tan50+tan20

tan70-cot20tan50tan20=tan50+tan20

tan70-(cot20tan20)tan50=tan50+tan20

tan70-(1)tan50=tan50+tan20

tan70-tan50=tan50+tan20

tan70=tan20+2tan50

LHS = RHS

Hope this helps!

Answer 2

Already Answered this Question So Copy Pasted It.

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${ \large{ \underline{ \sf{Solution :}}}}$

To Prove :

${ \footnotesize{ \rm{2tan50° = tan70°-tan20°}}}$

As We Know,

$\blue\star\boxed{ \displaystyle \rm{ \tan(A - B) = \frac{ \tan \: A - \tan \: B }{1 + \tan \:A \:\tan\: B } }}$

$\red\star\boxed{ \displaystyle \rm{ \tan(A+B) = \frac{ \tan \: A - \tan \: B }{1 - \tan \:A \:\tan\:B } }}$

Now,

$\displaystyle \rm{ \implies \: \tan(50^{0} ) = \tan(70 ^{0} - 20^{0}) }$

${\implies\displaystyle \rm{ = \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{1 + \tan \: 70 ^{0} \tan \: 20 {}^{0} } }}$

${\implies\displaystyle \rm{ \tan \:( 50 {}^{0} ) = \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{1 + \tan \: 70 ^{0} \tan \: 20 {}^{0} }...(1) }}$

$\displaystyle \rm{ \implies \: \tan(90 ^{0}) }$

$\displaystyle \rm{ \implies \: \tan(70 ^{0}) + (20^{0} )}$

${\implies\displaystyle \rm{= \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{1 - \tan \: 70 ^{0} \tan \: 20 {}^{0} }}}$

$\displaystyle \rm{ \implies \: \tan(90 ^{0}) }$

${\implies\displaystyle \rm{= \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{1 - \tan \: 70 ^{0} \tan \: 20 {}^{0} }}}$

Thus,

${ \implies \displaystyle \rm{1 - \tan \: 70 ^{0} \tan \:20 ^{0} = 0 }}$

${ \implies \displaystyle \rm{ \tan \: 70^{0} \tan \: 20 ^{0} = 1}}$

Subsitute This Result in (1)

★We Get,

${\implies\displaystyle \rm{ \tan \:( 50 {}^{0} ) = \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{1 + 1 }}}$

${\implies\displaystyle \rm{ \tan \:( 50 {}^{0} ) = \dfrac{ \tan \: 70 ^{0} -20 {}^{0} }{2}}} \:$

$\displaystyle \rm{ \tan \: 70 {}^{0} - \:tan \: 20 {}^{0} = 2 \tan \: 50 {}^{0} }$

Hence Proved.

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