Question

solve this question....

give proper answer ☺

Answer 1

Yeah sure for your help

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Nice Question and also hard .......

Here is your answer which you are searching for

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For the first answer refer to this attachment with my answer

use the formula

sinasinb = -1/2 (cos ( a + b) - Cos ( a-b) )

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Second Answer

Let y= tanA tanB =tanA tan (Pie/3 -A)

tan pie /3 -tanA
tanA _____________
1+ tan pie/3 tanA

√3 -tanA
==> tanA. _______
1+√3tanA

since, y + √3 tan A . y =√3 tanA -tan^2A

==> tan^2A + √3 (y-1) tanA +y = 0

since tanA is real

Discriminate >_ 0 ==> 3 (y -1 )^2 -4y >_ 0

==> 3 ( y^2 -2y +1 ) -4y >_ 0

==> 3y^2 -10y +3 >_ 0

==> ( 3y -1 ) (y -3 ) >_ 0

==> y <_ 1/3 or y >_ 3

But each of tanA , tanB is less than √3

[ if A =0 ,B = pie /3 and if B = 0 ,A =pie/3 ]

.: y >_ 3 is not possible. :. y <_ 1/3

===> tanA tanB <_3

========> Maximum value of tanA tanB = 1/3

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@Brainlestuser

Answer 2

The answer is explained in the attachment.

(1) Option (A)

(2) Option (B).

Hope this helps!