Question

a+b+c=90 solve tanA×tanB+tanB×tanC+tanC×tanA=1​

Answer 1

See the answer Below Mate:-

=>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

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

Answer 2

Answer:

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