Question

8. Prove that: (a) 2tan50° = tan70° - tan20°​

Answer 1

Solution:

To Prove:

$\rm \longrightarrow 2\tan(50^{ \circ} ) = \tan( {70}^{ \circ} ) - \tan( {20}^{ \circ} )$

Talking Left Hand Side, we get:

$\rm = 2\tan(50^{ \circ})$

Can be written as:

$\rm = 2\tan(70^{ \circ} - {20}^{ \circ} )$

We know that:

$\bigstar \underline{ \boxed{ \rm \tan( \alpha - \beta ) = \dfrac{ \tan( \alpha ) - \tan( \beta ) }{1 + \tan( \alpha ) \cdot \tan( \beta ) }}}$

Using this result, we get:

$\rm = 2 \times \dfrac{\tan({70}^{\circ})-\tan({20}^{\circ})}{1 + \tan({70}^{\circ})\tan({20}^{\circ}) }$

$\rm = 2 \times \dfrac{\tan({70}^{\circ})-\tan({20}^{\circ})}{1 + \tan({90}^{\circ} - {20}^{ \circ} )\tan({20}^{\circ}) }$

$\rm = 2 \times \dfrac{\tan({70}^{\circ})-\tan({20}^{\circ})}{1 + \cot({20}^{ \circ} )\tan({20}^{\circ}) }$

$\rm = 2 \times \dfrac{\tan({70}^{\circ})-\tan({20}^{\circ})}{1 + 1}$

$\rm = 2 \times \dfrac{\tan({70}^{\circ})-\tan({20}^{\circ})}{2}$

$\rm = \tan({70}^{\circ})-\tan({20}^{\circ})$

Taking Right Hand Side, we get:

$\rm = \tan({70}^{\circ})-\tan({20}^{\circ})$

We observe that LHS = RHS (Hence Proved)

Learn More:

Sum/Difference Identities.

$\rm1. \: \sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$

$\rm2.\: \cos(\alpha+\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$

$\rm 3.\: \tan(\alpha+\beta) = \dfrac{ \tan( \alpha) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }$

$\rm 4. \: \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$

$\rm 5.\: \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$

$\rm 6.\: \tan(\alpha - \beta) = \dfrac{ \tan( \alpha) - \tan( \beta ) }{1 + \tan( \alpha ) \tan( \beta ) }$

Answer 2

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

To Prove :

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

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,

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

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

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

Hence Proved.

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Additional Information :

$\bigstar \: \underline{ \boxed{ \rm\dfrac{d}{dx}( {x}^{n} ) =n {x}^{n - 1} }}$

$\bigstar \:\underline{ \boxed{ \rm\dfrac{d}{dx}( C ) =0}}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(C \times f) = C\dfrac{d}{dx}f}}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(f \pm g) = \dfrac{d}{dx}f \pm \dfrac{d}{dx}g}}$

$\bigstar \boxed{\displaystyle \rm \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} + C}$

$\bigstar \: \underline{ \boxed{ \rm \dfrac{d}{dx}(fg) =f \dfrac{d}{dx}g + g \dfrac{d}{dx}f }}$