How to prove sin(θ+ϕ)cos(θϕ)=tanθ+tanϕ1+tanθtanϕ?

3 Answers
May 26, 2018

Please see the proof below

Explanation:

We need

sin(a+b)=sinacosb+sinbcosa

cos(ab)=cosacosb+sinasinb

Therefore,

LHS=sin(θ+ϕ)cos(θϕ)

=sinθcosϕ+cosθsinϕcosθcosϕ+sinθsinϕ

Dividing by all the terms bycosθcosϕ

=sinθcosϕcosθcosϕ+cosθsinϕcosθcosϕcosθcosϕcosθcosϕ+sinθsinϕcosθcosϕ

=sinθcosθ+sinϕcosϕ1+sinθcosθsinϕcosϕ

=tanθ+tanϕ1+tanθtanϕ

=RHS

QED

May 26, 2018

See Explanation

Explanation:

Let
y=sin(θ+ϕ)cos(θϕ)

y=sinθcosϕ+cosθsinϕcosθcosϕ+sinθsinϕ

Dividing by cosθ,

y=tanθcosϕ+sinϕcosϕ+tanθsinϕ

Dividing by cosϕ,

y=tanθ+tanϕ1+tanθtanϕ

hence proved.

May 26, 2018

see explanation

Explanation:

using the trigonometric identities

xsin(x+y)=sinxcosy+cosxsiny

xcos(xy)=cosxcosy+sinxsiny

consider the left side

=sinθcosϕ+cosθsinϕcosθcosϕ+sinθsinϕ

divide terms on numerator/denominator by cosθcosϕ
and cancel common factors

=sinθcosϕcosθcosϕ+cosθsinϕcosθcosϕcosθcosϕcosθcosϕ+sinθsinϕcosθcosϕ=sinθcosθ+sinϕcosϕ1+sinθcosθ×sinϕcosϕ

=tanθ+tanϕ1+tanθtanϕ

=right side verified