Question

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Q. Prove that
$\tan \alpha (1 + \sec2 \alpha ) (1 + \sec4 \alpha ) = \tan(4 \alpha )$
​

Answer 1

Answer:

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Answer 2

$\tan \alpha (1 + \sec2 \alpha ) (1 + \sec4 \alpha ) = \tan(4 \alpha )$

___________ [ GIVEN ]

We have to prove that L.H.S. = R.H.S.

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Taking L.H.S.

i.e.

=> $\tan \alpha (1 + \sec2 \alpha ) (1 + \sec4 \alpha )$

We know that

secA = 1/cosA

=> $\tan \alpha (1 + \dfrac{1}{ \cos2 \alpha }) (1 + \dfrac{1}{ \cos4 \alpha} )$

=> $\tan \alpha (\dfrac{cos2 \alpha \: + \: 1}{ \cos2 \alpha }) ( \dfrac{cos4 \alpha \: + \: 1}{ \cos4 \alpha} )$

Cos2A = 2 cos²A - 1

=> $\tan \alpha (\dfrac{2cos^{2} \alpha \: - \: 1 \: + \: 1}{ \cos2 \alpha }) ( \dfrac{cos4 \alpha \: + \: 1}{ \cos4 \alpha} )$

=> $\tan \alpha (\dfrac{2cos^{2} \alpha}{ \cos2 \alpha }) ( \dfrac{cos4 \alpha \: + \: 1}{ \cos4 \alpha} )$

=> $\tan \alpha (\dfrac{2cos^{2} \alpha \: - }{ \cos2 \alpha }) ( \dfrac{2cos^{2} 2 \alpha \: - \: 1 \: + \: 1}{ \cos4 \alpha} )$

=> $\tan \alpha (\dfrac{2cos^{2} \alpha }{ \cos2 \alpha }) ( \dfrac{2cos^{2} 2 \alpha}{ \cos4 \alpha} )$

=> $\dfrac{sin \alpha}{cos\alpha} (\dfrac{2cos^{2} \alpha }{ \cos2 \alpha }) ( \dfrac{2cos^{2} 2 \alpha}{ \cos4 \alpha} )$

=> $sin \alpha \: (\dfrac{2cos\alpha }{ \cos2 \alpha }) \: ( \dfrac{2cos^{2} 2 \alpha}{ \cos4 \alpha} )$

=> $sin \alpha \: (2cos\alpha) \: ( \dfrac{2cos 2 \alpha}{ \cos4 \alpha} )$

=> $( \dfrac{2sin \alpha \: cos \alpha( 2cos 2 \alpha)}{ \cos4 \alpha} )$

=> $( \dfrac{sin 2\alpha \:( 2cos 2 \alpha)}{ \cos4 \alpha} )$

=> $( \dfrac{2sin 2\alpha \:cos 2 \alpha}{ \cos4 \alpha} )$

tan4A = 2sin2A cos2A

=> $( \dfrac{sin 4\alpha}{ \cos4 \alpha} )$

=> $tan 4\alpha$

L.H.S. = R.H.S.

______ [ HENCE PROVED ]

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