Question

Please give me the answer....

Answer 1

prove  tan(a+b)=(tan a+tanb)/(1-tana*tanb)


Here, we use the relations that tan x = sin x / cos x.

sin (a + b) = sin a * cos b + cos a * sin b

cos ( a+ b) = cos a * cos b - sin a * sin b

tan (a + b) = sin (a + b) / cos ( a + b)

=> [sin a * cos b + cos a * sin b] / [cos a * cos b - sin a * sin b]

divide all the terms by cos a * cos b

=>[ (sin a * cos b)/(cos a * cos b)+ (cos a * sin b)/(cos a * cos b)] / [(cos a * cos b)/(cos a * cos b) - (sin a * sin b)/(cos a * cos b)]

=> [(sin a / cos a) + (sin b / cos b)]/[ 1 - (sin a / cos a)*( sin b/ cos b)]

=> (tan a + tan b) / ( 1 - tan a * tan b)

Therefore we have tan (a + b) = (tan a + tan b) / ( 1 - tan a * tan b)

Answer 2

sin A=1/✓2. ,cos B =√3/2

cos A=√(1-sin^2A)
=✓(1-1/2)
=1/√2

sin B =√(1-cos^2B)
=√(1-3/4)=1/2


tanA=sinA/cosA=(1/√2)×(√2/1)=1
tan B=sin B/cosB =(1/2)×2/√3=1/√3


now,. tanA+tanB=(√3+1)/√3
and,, 1-tanA.tanB=1-1×1/√3=(√3-1)/√3


Now the answer is
=(√3+1)√3÷(√3-1)/√3
=(√3+1)/(√3-1)
=(√3+1)^2/2
=2+√3
this is your answer