If tanα=(m/m+1), tanβ=(1/2m+1), then α+β is equal to

Question

If tanα=(m/m+1), tanβ=(1/2m+1), then α+β is equal to (A) (π/2) (B) (π/3) (C) (π/6) (D) (π/4)

Tardigrade Admin · Accepted Answer

Given tanα=(m/m+1), tanβ=(1/2m+1) ∴ tan(α+β)=(tan α+tan β/1-tan α tan β) =((m/m+1)+(1/2m+1)/1-(m)m+1×(2)2m+1 =(2m2+m+m+1/(m+1)(2m+1)-m) =(2m2+2m+1/2m2+2m+1)=1 ⇒ tan(α+β)=1 ⇒ α+β=(π/4)

Q. If $t an α = \frac{m}{m + 1}$ , $t an β = \frac{1}{2 m + 1}$ , then $α + β$ is equal to

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{4}$

Q. If tanα=m+1m​, tanβ=2m+11​, then α+β is equal to

2π​

3π​

6π​

4π​

Given tanα=m+1m​, tanβ=2m+11​ ∴tan(α+β)=1−tanαtanβtanα+tanβ​ =1−m+1m​×2m+12​m+1m​+2m+11​​ =(m+1)(2m+1)−m2m2+m+m+1​ =2m2+2m+12m2+2m+1​=1 ⇒tan(α+β)=1 ⇒α+β=4π​

Q. If $t an α = \frac{m}{m + 1}$ , $t an β = \frac{1}{2 m + 1}$ , then $α + β$ is equal to

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{4}$