# Calculate tg( a + b ) if a belongs to ( 0 , pi/2 ) and b belongs to ( pi/2 , pi ) . sin a = 1/2 ; sin b = 2/3

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WE know that:

tg(a+b)=(tg(a)+tg(b))/(1-tg(a)tg(b)).....(1)

Now let us calculate:

cos b= -sqrt(1-sin^2 b) = -sqrt(1-4/9)=-sqrt(5/9)

tg b= (2/3)/-(sqrt(5/9)= -2/sqrt(5)=-2sqrt(5)/5

cos a= sqrt(1-sin^2 a) = (1-(1/4))^1/2= sqrt(3/4)

tg a= (1/2)/sqrt(3/4)= 1/sqrt(3)=sqrt(3)/3

Now substitute tg a, tg b in equation (1):

==> tg (a+b)= [sqrt3/3)-2sqrt(5)/5]/[1+2(sqrt3/3)*(sqrt5/5)

= (5sqrt3-6sqrt5)/(15+2sqrt15)

To calculate tg(a+b), where a is in (0,pi/2) and bis in (pi/2, pi)

and sina = 1/2, sinb = 2/3.

solution:

tg (a+b) = (tana+tanb)/(1-tana+tanb),.......eq(1)

sina = 1/2. g a = sina/sqrt(1-sin^2a) = (1/2)/sqrt(1-1/2^2) = 1/sqr13, as a is in (0,pi/2)

tgb = - sinb/ (1-sin^2b) , - sign is because tan is -ve in 2nd quadrant or in (pi/2 , pi)

= -(2/3)/ sqrt[1-2/3)^2] = 2/sqrt5. Substituting the values of tag a and tg b in eq(1) we get:

tan(a+b) = (1/sqrt3 -2/sqrt5)/{ 1- sqrt(1/3)(-2/5)} multiplying both numerator and denominator by sqrt15.

= (sqrt5- 2sqrt3)/[(sqrt15) +1]

=( sqrt5-2sqrt3)[(sqrt(15)-1]/14 , Rationalise the denominator:

= (sqrt75-2sqrt45+sqrt5-2sqrt3)/14

= (5sqrt3 - 6sqrt5+sqrt5-2sqrt3)/14

=(3sqrt3-5sqrt5)/14 = -0.427441961

First, we need to establish the sign of tg a and tg b. Based on the facts from enunciation, tg a belongs to the first quadrant and it is positive and tg b belongs to the second quadrant and it is negative.

tg a = sina a/cos a

cos a = sqrt( 1 - sin^2a )

cos a = sqrt( 1 - 1/4 ) = sqrt 3 /2

cos b = - sqrt(1-4/9 ) = - sqrt5/3

tg a = ( 1/2 )/sqrt 3/2

tg b = ( 2/3 )/[ -sqrt5 ]/3 = - 2sqrt5 ]/3

tg ( a + b ) = ( tg a + tg b )/( 1 - tga*tgb )

tg ( a + b ) = { ( 1/2 )/sqrt 3 /2 + ( - 2sqrt5 )/3 }/{ 1 - ( 1/2 )/sqrt 3 ]/2*[ - 2sqrt5 ]/3 }

tg ( a + b ) = ( 5sqrt 3 - 6 sqrt5 )/( 15 + 2sqrt15 )