Question

tan invese 1+tan inverse 2+tan inverse 3

Answer 1

I hope this is the correct answer.

Answer 2

hello users .........

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we have to find :
$tan ^{-1}1 + tan^{-1}2 + tan^{-1} 3 = ?$

solution :-
we know that:
$tan^{-1}x +tan^{-1}y +tan^{-1}z = tan^{-1}( \frac{x + y +z - xyz}{1 - xy - yz - zx})$

here,
$tan ^{-1}1 + tan^{-1}2 + tan^{-1} 3$

= $tan ^{-1}[ \frac{( 1+2 +3 - 1*2*3)}{(1 - 1*2 - 2*3 - 3*1)} ]$

=$tan^{-1} [ \frac{(6-6)}{(-10)} ]$

=$tan^{-1} (0)$

= 0 answer    .....( because $tan^{-1} (0)$ = 0 )

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✪✪ hope it helps ✪✪