Question

if sin(a+b)=n sin (a-b) and (n≠-1) then prove that cot a=n-1/n+1 cot-b​

Answer 1

Step-by-step explanation:

sinθ=nsin(θ+2α)

sin(θ+2α)

sinθ

=n

apply componendo dividendo rule,

sinθ−sin(θ+2α)

sinθ+sin(θ+2α)

=

n−1

n+1

apply sinC+sinD and sinC−sinD

2sinαcos(θ+α)

2sin(θ+α)cosα

=

n−1

n+1

tanα

tan(θ+α)

=

n−1

n+1

tan(θ+α)=

n−1

n+1

tanα