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

Question

If tan θ =3tan ⁡ 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 =3tan â¡ φ tan (θ - φ)=(tan â¡ θ - tan â¡ φ/1 + tan â¡ θ tan â¡ φ) =(2 tan φ/1 + 3 tan2 â¡ φ)=(2/cot â¡ φ + 3 tan â¡ φ) Using AM≥ GM (cot φ + 3 tan â¡ φ/2)≥ √3 ⇒ (2/cot φ + 3 tan â¡ φ)≤ (1/√3) ⇒ tan (θ - φ)≤ (1/√3) ⇒ tan 2(θ-φ) ≤ (1/3)

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

$1$

$\frac{1}{3}$

$2$

$4$

Q. If tanθ=3tan⁡ϕ , then the maximum value of (tan)2(θ−ϕ) is (where, tanϕ>0 )

1

31​

2

4

tanθ=3tan⁡ϕ tan(θ−ϕ)=1+tan⁡θtan⁡ϕtan⁡θ−tan⁡ϕ​ =1+3tan2⁡ϕ2tanϕ​=cot⁡ϕ+3tan⁡ϕ2​ Using AM≥GM 2cotϕ+3tan⁡ϕ​≥3​ ⇒cotϕ+3tan⁡ϕ2​≤3​1​ ⇒tan(θ−ϕ)≤3​1​ ⇒tan2(θ−ϕ)≤31​

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

$1$

$\frac{1}{3}$

$2$

$4$