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Prove tan(a+b)

Q 5 (a) prove that     tan(A+B) =    tan A+tan B
__________
1- tan A tan B

(b) show that value of 75 degrees = 2+root3

(c) is hence or otherwise show tan(15 degrees) in surd form

I know how to do part a its only a proof but could you show me how to do (b) and (c) thanks

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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Prove : Tan ( A+ B ) = (Tan A + Tan B ) / ( 1 - Tan A Tan B) :

Tan ( A + B ) = Sin ( A + B) / Cos ( A +B)

( Sin A Cos B + Sin B Cos A ) / ( Cos A Co B - Sin A Sin B)  ( 1)

Divide both sides of equation ( 1) by Cos A . Cos  B

( Sin A Cos B + Sin B Cos A )/ (CosA Cos  B )  / right side of ( 1)

Tan A + Tan B ( i)

Cos A Cos B - Sin A Sin B / Cos A . Cos B =

( 1 - Tan A Tan B )  ( ii )

Therefore:

Tan ( A + B ) =  ( i) / ( ii )

= ( Tan A + Tan B )/ ( 1 - Tan A Tan B)

Tan ( 75) = (Tan 45° + Tan 30°) / ( 1 - Tan45°. Tan30°)

( 1 + √3 /3 ) /( 1 - √3/ 3) = 2 + √3

Tan 15 = ( Tan 45° - Tan30° ) / (1 + √3/ 3 )

= ( 1 - √3 / 3 )/ (1  + √3 /3 ) =

=  ( 3 - √3 ) / ( 3 + √3 )

= 2 - √3
Shelly J. | Excellent Maths Tutoring for academic successExcellent Maths Tutoring for academic su...
5.0 5.0 (274 lesson ratings) (274)
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Hi Nick,

(a)   tan(A+B)=sin(A+B)/cos(A+B)
=sinAcosB+cosAsinB
cosAcosB-sinAsinB

divide the numerator and denominator by cosAcosB

=sinA/cosA+sinB/cosB
1-sinAsinB/cosAcosB
=tanA+tanB
1-tanAtanB

(b)   tan75=tan(30+45)
=tan30+tan45
1-tan30tan45

=1/√3+1
1-1/√3

=(1+√3)/(√3-1)

multiply the numerator and denominator by √3+1

=(1+√3)²/(√3)²-(1)²

=(1+2√3+3)/(3-1)

=(4+2√3)/2
=2+√3

(c)  tan15=tan(60-45)
=tan60-tan45
1+tan60tan45

=(√3-1)/(1+√3)

multiply the numerator and denominator by 1-√3

and √3-1=-(1-√3)

=-(1-√3)²/(1)²-(√3)²

=-(1-2√3+3)/-2

=(4-2√3)/2

=2-√3