Question

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If A+B+C=90, Then prove that
tanA tanB + tanB tanC + tanC tanA = 1

Explain in detail

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Answer 1

HELLO DEAR,

we know that:-

$tan(a + b) = \frac{tana + tanb}{1 - tana.tanb}$

given that:-

A + B + C = 90°

=> A + B = 90° - C

multiply both side by tan

=> tan(A + B) = tan(90° - C )

=> tan(A + B) = cotC

$\frac{tana + tanb}{1 - tana \times tanb} = \frac{1}{tanc} \\ \\ = > tanc(tana + tanb) = 1 - tana.tanb \\ \\ = > tana.tanc + tanb.tanc = 1 - tana.tanb \\ \\ = > tana.tanc + tanb.tanc + tana.tanb = 1$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hey mate.
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The solution is in the pic.

Hope it helps !!