If tan- m - m -1- tan-1-2 m -1- then - n - Z A- n -2- n -3- n -4D- n -4

The correct option is D tan (α+β) = tan α + tan β/1 tan α tan β = m/m+1+1/2m+1/1 m/m+1.1/2m+1=2m^2 + 2m+1/2m^2+2m+1=1 ⇒α + β = n π + π/4, n ϵ Z. ...

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Byju's Answer

Standard XII

Physics

Introduction

If tanα= m / ...

Question

$I f t a n α = \frac{m}{(m + 1)}, t a n β = \frac{1}{(2 m + 1)}, t h e n α + β = (n ϵ Z)$

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Solution

The correct option is D
$t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β} = \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} . \frac{1}{2 m + 1}} = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1 \Rightarrow α + β = n π + \frac{π}{4}, n ϵ Z .$

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If tan α=m(m+1),tan β=1(2m+1),thenα+β=(nϵZ)

The correct option is D tan (α+β)=tan α+tan β1−tan α tan β=mm+1+12m+11−mm+1.12m+1=2m2+2m+12m2+2m+1=1⇒α+β=nπ+π4,nϵZ.

$I f t a n α = \frac{m}{(m + 1)}, t a n β = \frac{1}{(2 m + 1)}, t h e n α + β = (n ϵ Z)$

The correct option is D
$t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β} = \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} . \frac{1}{2 m + 1}} = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1 \Rightarrow α + β = n π + \frac{π}{4}, n ϵ Z .$