If α+β=(π/2) and β+γ, then tan α equals

Question

If α+β=(π/2) and β+&gamma;, then tan α equals (A) 2 (tanβ + tan&gamma;) (B) tanβ + tan&gamma; (C) tanβ + 2tan&gamma; (D) 2tanβ + tan&gamma;

Tardigrade Admin · Accepted Answer

Given, α+β=π/2 ⇒ α=(π/2)=β ⇒ tan α=tan(π/2-β) ⇒ tan α=cot β ⇒ tan α tan β=1 Again, β+gamma=α [given] ⇒ &gamma;=(α-β) ⇒ tan &gamma;=tan (α-β ⇒ tan &gamma;=(tan α-tanβ/1+tan α tanβ) ⇒ (tan α-tanβ/1+1) ∴ 2 tan &gamma; = tan α - tan β ⇒ tan α=tan β+2 tan &gamma;

Q. If $α + β = \frac{π}{2} an d β + γ$ , then $t an α$ equals

$2 (t an β + t anγ)$

$t an β + t anγ$

$t an β + 2 t anγ$

$2 t an β + t anγ$

Q. If α+β=2π​andβ+γ, then tanα equals

2(tanβ+tanγ)

tanβ+tanγ

tanβ+2tanγ

2tanβ+tanγ

Given, α+β=π/2⇒α=(π/2)=β ⇒tanα=tan(π/2−β) ⇒tanα=cotβ⇒tanαtanβ=1 Again, β+gamma=α[given] ⇒γ=(α−β) ⇒tanγ=tan(α−β ⇒tanγ=1+tanαtanβtanα−tanβ​ ⇒1+1tanα−tanβ​ ∴2tanγ=tanα−tanβ ⇒tanα=tanβ+2tanγ

Q. If $α + β = \frac{π}{2} an d β + γ$ , then $t an α$ equals

$2 (t an β + t anγ)$

$t an β + t anγ$

$t an β + 2 t anγ$

$2 t an β + t anγ$