If cos (θ +φ )=m cos (θ -φ ), then tan θ is equal to:

Question

If cos (θ +φ )=m cos (θ -φ ), then tan θ is equal to: (A)  [(1+m)/(1-m)] tan φ  (B)  [(1-m)/(1+m)] tan φ  (C)  [(1-m)/(1+m)]cotφ  (D)  [(1+m)/(1-m)]secφ

Tardigrade Admin · Accepted Answer

We have cos (θ+φ)=m cos(θ-φ) ⇒ cos θ cos φ- sin θ sin φ=m( cos θ cos φ+ sin θ sin φ) ⇒ cos θ cos φ(1-m)= sin θ sin φ(1+m) ⇒ tan θ=((1-m/1+m)) cot φ

Q. If $cos (θ + ϕ) = m cos (θ - ϕ),$ then $tan θ$ is equal to:

$[(1 + m) / (1 - m)] tan ϕ$

$[(1 - m) / (1 + m)] tan ϕ$

$[(1 - m) / (1 + m)] co tϕ$

$[(1 + m) / (1 - m)] sec ϕ$

We have $cos (θ + ϕ) = m cos (θ - ϕ)$
$\Rightarrow cos θ cos ϕ - sin θ sin ϕ = m (cos θ cos ϕ + sin θ sin ϕ)$
$\Rightarrow cos θ cos ϕ (1 - m) = sin θ sin ϕ (1 + m)$
$\Rightarrow tan θ = (\frac{1 - m}{1 + m}) cot ϕ$

Q. If cos(θ+ϕ)=mcos(θ−ϕ), then tanθ is equal to:

[(1+m)/(1−m)]tanϕ

[(1−m)/(1+m)]tanϕ

[(1−m)/(1+m)]cotϕ

[(1+m)/(1−m)]secϕ