Question

○●○ No. 36 Please ○●○

Answer 1

Thanka for the QN +_+

# We know :

----> tan ( 45° ) = 1

Now,

$\tan(A+B+C) = \tan(45 {}^{o} ) = 1$

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

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

$= > \sum \tan(A) - \prod \tan(A) = 1 - \sum \tan(A) \tan(B)$

$= > \sum( \tan(A) + \tan(A) \tan(B) ) = 1 + \prod \tan(A)$

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Or wait ^^"

$A+B=45 {}^{o}-C )$

$\tan(A+B) = \tan(45 {}^{o} -C)$

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

$= > \sum \tan(A) - \prod \tan(A) = 1 - \sum \tan(A) \tan(B)$

$= > \sum( \tan(A) + \tan(A) \tan(B) ) = 1 + \prod \tan(A)$
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Have a great day Allys ^_^

*( Both ways are exactly same ! Depends on which one's shorter ^^" )