Question

If tan A = 5/13 , π < A < 3π/2 and cos B = 3/5 , 3π/2 < B < 2π then cos (A - B) ? ​

Answer 1

Answer:

Given : sin A= -5/13 π<A<3π/2

cos B=3/5 3π/2< B<2π

To Find : Sin ( A + B)

Solution:

Sin ( A + B) = SinA CosB + cosA SinB

sin A= -5/13 π<A<3π/2

sin²A + cos²A = 1

=> cosA = - 12/13 as π<A<3π/2 hence cos is -ve

cos B=3/5 3π/2< B<2π

=> sin B = - 4/5 as 3π/2< B<2π hence sin is -ve

Sin ( A + B) = SinA CosB + cosA SinB

=> Sin ( A + B) = (-5/13) (3/5) + (-12/13) ( -4/5)

=> Sin ( A + B) = -3/13 + 48/65

=> Sin ( A + B) = 33/65

=> Sin ( A + B) = 0.5077

Answer 2

""" ❤️ Answer ❤️ """

Given, sin A = -513

We know that,

cos2A = 1 – sin2A = 1-(-513)2

= 1-25169

= 144169

∴ cos A = ±1213

Since,

A

π<A<3π2

∴ ‘A’ lies in the 3rd quadrant

∴ cos A < 0

∴ cos A = -1213

Also, cos B = 35

∴ sin2B = 1 – cos2B = 1-(35)2

= 1-925

= 1625

∴ sin B = ±45

Since,

B

3π2<B<2π

∴ ‘B’ lies in the 4th quadrant.

∴ sin B < 0

∴ sin B = -45

tan A =

A

A

sinAcosA=(-513)(-1213)=512

tan B =

B

B

sinBcosB=(-45)(35)=-43

tan (A + B) =

AB

AB

tanA+tanB1-tanAtanB

=

512-431-(512)(-43)

= (-3336)(5636)

= -3356

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