Question

If tan theta =a/b , then asintheta+bcostheta/asintheta-bcostheta is equal to

Answer 1

i m taking theta = € for convenience.
sin € / cos € = a/b ..
implies that sin € = (acos €)/b
put this value of sin € in question asked ...
you will get (a^2cos € + b^2 cos € )/a^2cos €-b^2cos € ...
taking cos € common.....
Ans comes to be a^2+b^2/a^2-b^2
...hope it helps ....
mark as....you know what i mean ...

Answer 2

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)

=rhs