Question

in triangle ABC,right-angle at B,if tanA=1/√3,find the value of
sina cosc+cosa sinc
cosa cosc-sina sinc

Answer 1

Answer:

sinA.cosC + cosA.sinC = 1

cosA.cosC - sinA.sinC = 0

Step-by-step explanation:

Formula used:

sin(A+B) = sinA.cosB + cosA.sinB

cos(A+B)= cosA. cosB - sinA.sinB

Given:

tanA=1/√3

A=30°

since B is right angle, C=60°

Now,

sinA.cosC + cosA.sinC

=sin(A+C)

=sin(30°+60°)

=sin90°

=1

cosA.cosC - sinA.sinC

=cos(A+C)

=cos(30°+60°)

=cos90°

=0

Answer 2

Answer:

SinACosC + CosASinC = 0

CosACosC - SinASinC = 0

Solution is independent of tanA value

Step-by-step explanation:

In triangle ABC,right-angle at B,if tanA=1/√3,find the value of

sina cosc+cosa sinc

cosa cosc-sina sinc

Its a right angle triangle at B

so ∠A + ∠C = 90

∠A = ∠90 - ∠C

Sin∠A = Sin(∠90 - ∠C) = CosC

Cos∠A = Cos(∠90 - ∠C) = SinC

SinACosC + CosASinC

= SinASinA + CosACosA

=Sin²A + Cos²A

= 1

CosACosC - SinASinC

= SinCCosC - CosCSInC

= SinCCosC - SinCCosC

= 0