Question

if sec tita =m and tan tita= n , then 1/m[(m+n)+1/m+n] is equal to

Answer 1

Answer:

cos(θ−ϕ)

cos(θ+ϕ)

=m

Apply componendo and dividendo

cos(θ+ϕ)−cos(θ−ϕ)

cos(θ+ϕ)+cos(θ−ϕ)

=

m−1

m+1

−(cos(θ−ϕ)−cos(θ+ϕ))

cos(θ+ϕ)+cos(θ−ϕ)

=

m−1

m+1

−2sin(θ)⋅sin(ϕ)

2cos(θ)⋅cos(ϕ)

=

m−1

1+m

2sin(θ)⋅sin(ϕ)

2cos(θ)⋅cos(ϕ)

=

1−m

1+m

cotθ⋅cotϕ=

1−m

1+m

tanθ=(

1+m

1−m

)⋅cotϕ

Step-by-step explanation:

hope it help you.......