Question

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Date
Prove tan 2 alpha - tan alpha = tan alpha sec2 alpha​

Answer 1

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:tan2 \alpha - tan \alpha = tan \alpha sec2 \alpha$

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

Consider LHS

$\rm :\longmapsto\:tan2 \alpha - tan \alpha$

We know

$\boxed{ \rm{ tanx = \frac{sinx}{cosx}}}$

So, using this, we get

$\rm \: = \: \: \dfrac{sin2 \alpha }{cos2\alpha } - \dfrac{sin\alpha }{cos\alpha }$

$\rm \: = \: \: \dfrac{sin2\alpha \: cos\alpha - sin\alpha \: cos2\alpha }{cos2\alpha \: cos\alpha }$

We know that

$\boxed{ \rm{ sin(x - y) = sinxcosy - sinycosx}}$

So, using this identity, we get

$\rm \: = \: \: \dfrac{sin(2\alpha \: - \: \alpha )}{cos2\alpha \: cos\alpha }$

$\rm \: = \: \: \dfrac{sin\alpha }{cos2\alpha \: cos\alpha }$

$\rm \: = \: \: \dfrac{1}{cos2\alpha } \times \dfrac{sin\alpha }{cos\alpha }$

$\rm \: = \: \: sec2\alpha \: tan\alpha$

Hence, Proved

Additional Information :-

$\boxed{ \rm{ sin(x + y) = sinxcosy +sinycosx}}$

$\boxed{ \rm{ cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \rm{ cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \rm{ tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \rm{ tan(x - y) = \frac{tanx - tan y}{1 + tanxtany}}}$

$\boxed{ \rm{ {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y)}}$

$\boxed{ \rm{ {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y)}}$

Answer 2

$\fbox{ tan2α - tanα }$

We know that :

$\large\rm{ tanx = \large\frac{sinx}{cosx} }$

By Applying this :-

$= \frac{ \sin2 \alpha }{ \cos2 \alpha } - \frac{ \sin \alpha }{ \cos \alpha }$

$= \frac{ \sin2 \alpha \cos \alpha - \sin \alpha \cos2 \alpha }{ \cos2 \alpha \cos \alpha }$

We know that :

$\fbox{ sib(x-y) = sinxcosy-sinycosx }$

By using this we get :

$= \frac{ \sin(2 \alpha - \alpha ) }{ \cos2 \alpha \cos \alpha }$

$= \frac{ \sin \alpha }{ \cos2 \alpha \cos \alpha }$

$= \frac{1}{ \cos2 \alpha } \times \frac{ \sin \alpha }{ \cos \alpha }$

$= \sec2 \alpha \tan \alpha$

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