Sum and product are different things, silly
True or False?
If A , B and C are angles of a non-right triangle, then tan A + tan B + tan C = tan A × tan B × tan C .
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1 solution
You did a mistake here.
tan A ( 1 − tan B tan C ) = − ( tan B + tan C ) ⟹ − tan A = 1 − tan B tan C tan B + tan C ⟹ tan ( 1 8 0 ∘ − A ) = tan ( B + C )
But rest of your solution is nice. +1 by me.
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Ya, i skipped the step .. thanks
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Now
tan A ( 1 − tan B tan C ) = − ( tan C + tan B )
Hence − tan A = ( tan C + tan B ) ( 1 − tan B tan C )
You've written the RHS of the final equation above as product, whereas, it should be
Now
tan A ( 1 − tan B tan C ) = − ( tan C + tan B )
Hence − tan A = ( 1 − tan B tan C ) ( tan C + tan B )
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@Tapas Mazumdar – Bro.. its not Latexed...
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@Md Zuhair – Use the following LaTeX for it:
\dfrac{(\tan C + \tan B)}{(1- \tan C \tan B)}
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@Tapas Mazumdar – Ok. Thanks...
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@Md Zuhair – You're welcome brother. :)
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@Tapas Mazumdar – Please respond to my report. Am I missing something? Thanks, Ed Gray
Here we can see that t a n A + t a n B + t a n C = t a n A x t a n B x t a n C .
Now
t a n A ( 1 − t a n B t a n C ) = − ( t a n C + t a n B )
Hence − t a n A = ( 1 − t a n B t a n C ) ( t a n C + t a n B )
hence 1 8 0 − A = B + C hence A + B + C = 1 8 0 .
Hence A , B ,C Form a traingle. So answer is T R U E .