Question

if tan θ = a/b then find the value of (b sin θ-a cos θ)/ (b sin θ+a cos θ)

Answer 1

Answer:

Step-by-step explanation:

Tan θ = a/b

sinθ/cosθ = a/b------(1)

lhs = (a sinθ- b cos θ)/(a sinθ + b cosθ)

= (asinθ/cosθ - bcosθ/cosθ)/(asinθ/cosθ+ bcosθ/cosθ)

dividing each term with cosθ

=(asinθ/cosθ-b/1)/(asinθ/cosθ+b/1)

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

[from (1)]

= (a^2-b^2)/(a^2+b^2)

Answer 2

Answer:

Tan 0 = a/b

sine/cose = a/b -(1)

Ihs = (a sine- b cos 0)/(a sine + b cose)

= (asine/cose - bcose/cose)/(asine/cose+ bcose/cose)

dividing each term with cose

=(asine/cose-b/1)/(asine/cose+b/1)

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

[from (1)]

= (a^2-b^2)/(a^2+b^2)