If tan θ=3 tan φ, then the maximum value of tan 2(θ-φ) is (where, tan φ0)

Question

If tan θ=3 tan φ, then the maximum value of tan 2(θ-φ) is (where, tan φ>0) (A) 1 (B) (1/3) (C) 2 (D) 4

Tardigrade Admin · Accepted Answer

tan θ=3 tan φ tan (θ-φ)=( tan θ- tan φ/1+ tan θ tan φ) =(2 tan φ/1+3 tan 2 φ)=(2/ cot φ+3 tan φ) Using AM ≥ GM ( cot φ+3 tan φ/2) ≥ √3 ⇒ (2/ cot φ+3 tan φ) ≤ (1/√3) ⇒ tan (θ-φ) ≤ (1/√3) ⇒ tan 2(θ-φ) ≤ (1/3)

Q. If $tan θ = 3 tan ϕ$ , then the maximum value of $tan^{2} (θ - ϕ)$ is (where, $tan ϕ > 0)$

$1$

$\frac{1}{3}$

$2$

$4$

Q. If tanθ=3tanϕ, then the maximum value of tan2(θ−ϕ) is (where, tanϕ>0)

1

31​

2

4

tanθ=3tanϕ tan(θ−ϕ)=1+tanθtanϕtanθ−tanϕ​ =1+3tan2ϕ2tanϕ​=cotϕ+3tanϕ2​ Using AM ≥ GM 2cotϕ+3tanϕ​≥3​ ⇒cotϕ+3tanϕ2​≤3​1​ ⇒tan(θ−ϕ)≤3​1​ ⇒tan2(θ−ϕ)≤31​

Q. If $tan θ = 3 tan ϕ$ , then the maximum value of $tan^{2} (θ - ϕ)$ is (where, $tan ϕ > 0)$

$1$

$\frac{1}{3}$

$2$

$4$