Question

$tan \: 2 \alpha - \tan( \alpha ) = \tan( \alpha ) \sec(2 \alpha ) prove \: it$
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Answer 1

$\tan(2 \alpha ) - \tan( \alpha ) = \sin(2 \alpha - \alpha ) \div \cos(2 \alpha ) \cos( \alpha ) \\ = \sin( \alpha ) \div \cos( \alpha ) \cos(2 \alpha ) \\ = \tan( \alpha ) \div \cos(2 \alpha ) \\ = \tan( \alpha ) \sec(2 \alpha )$

Step-by-step explanation:

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Answer 2

Answer:

To prove : $\tan 2 \alpha - \tan \alpha =\tan\alpha .\sec 2\alpha$

Explanation:

$\tan 2 \alpha - \tan \alpha \\\\\tan 2 \alpha = \frac{2\tan alpha}{(1-\tan^2\alpha)}\\\\\frac{2\tan alpha}{(1-\tan^2\alpha)}-\tan\alpha\\\\\frac{2\tan\alpha}{1-\frac{\sin^2\alpha}{\cos^2\alpha}}-\tan\alpha\\\\\frac{2\tan \alpha \times \cos^2\alpha }{(\cos^2\alpha - \sin^2\alpha)}-\tan\alpha\\\\\frac{2\tan \alpha \cos^2\alpha}{\cos 2\alpha}-\tan \alpha\\\\2\tan \alpha\cos^2 \alpha\sec 2\alpha-\tan\alpha\\\\\tan\alpha(\frac{2\cos^2\alpha}{\cos 2 \alpha}-1)\\\\\tan\alpha(\frac{1}{\cos 2\alpha})$

$\\\\=\tan\alpha .\sec 2\alpha$

hence proved

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