Question

Please help me solve this

Answer 1

Answer:

See below

Step by step explanation

$\tan( \frac{\pi}{4} + A) \tan( \frac{3\pi}{4} + A) = - 1 $

L.H.S

$\tan( \frac{\pi}{4} + A) \tan( \frac{3\pi}{4} + A )$

We know that ,

$\tan(A + B) = \frac{tan \: A + tan \: B}{1 - \tan \: A \: \tan \: B }$

$( \frac{ \tan( \frac{\pi}{4} + \tan \: A ) }{1 - \tan \frac{\pi}{4} \tan \: A} ) \: (\frac{ \tan \frac{3\pi}{4} + \tan \: A}{1 - \tan \frac{3\pi}{4} \tan( \: A) } )$

$( \frac{1 + \tan \: A }{1 - \tan\: A} )( \frac{ \tan( x - \frac{x}{4} + \tan \: A ) }{1 - \tan(\pi - \frac{\pi}{4} ) \: \tan \: A })$ (tan π / 4 = 1 )

$( \frac{1 + \tan \: A}{ -1 - \: \tan \: A } )( \frac{ - \tan( \frac{\pi}{4} + \tan \: A ) }{1 - ( - \tan \: \frac{\pi}{4}) \: \tan \: A } )$ [ tan ( π - B ) = - tan∅ ]

$( \frac{1 + tan \: A}{1 - tan \: B} )( \frac{ - 1 + \tan\: A }{1 + \tan \: A } )$

$= \frac{ - (1 - \tan\: A)}{(1 - \tan \: A) }$

$= - 1$

L.H.S = R.H.S

Proved

Hope this helps..

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