If tan- m - m -1 and tan-1-2 m -1- then -IIT 1978- EAMCET 1992- Roorkee 1998- JMI EEE 2001-A- None of theseB- -4C- -3D- -6

The correct option is B We have, tan α = nbsp; m/m + 1 and tan β = nbsp; 1/2m + 1 We know tan(α + β) = nbsp; tan α + tanβ/1 tanα tanβ = nbsp; m/m+1+ ...

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Byju's Answer

Standard XII

Physics

Introduction

If tanα= m / ...

Question

If tan $α$ = $\frac{m}{m + 1}$ and tan $β$ = $\frac{1}{2 m + 1}$ , then $α$ + $β$ =

[IIT 1978; EAMCET 1992; Roorkee 1998; JMI EEE 2001]

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None of these

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Solution

The correct option is B

We have, tan $α$ = $\frac{m}{m + 1}$ and tan $β$ = $\frac{1}{2 m + 1}$

We know tan( $α + β)$ = $\frac{t a n α + t a n β}{1 - t a n α t a n β}$

= $\frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{(m + 1)} \frac{1}{(2 m + 1)}}$ = $\frac{2 m^{2} + m + m + 1}{2 m^{2} + m + 2 m + 1 - m}$

= $\frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1}$ = 1 $\Rightarrow$ tan( $α$ + $β$ ) = tan $\frac{π}{4}$

Hence, $α + β$ = $\frac{π}{4}$ .

Trick: As $α + β$ is independent of m, therefore put m = 1,

then tan $α$ = $\frac{1}{2}$ and tan $β$ = $\frac{1}{3}$ . Therefore,

tan $(α + β)$ = $\frac{(\frac{1}{2}) + \frac{1}{3}}{1 - (\frac{1}{6})}$ = 1. Hence $α$ + $β$ = $\frac{π}{4}$ .

(Also check for other values of m).

Suggest Corrections

Similar questions

tan α = \frac{m}{m + 1}

and

tan β = \frac{1}{2 m + 1}

then prove that

α + β = \frac{π}{4}

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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Find

a

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If tanα = mm+1 and tanβ = 12m+1, then α + β = [IIT 1978; EAMCET 1992; Roorkee 1998; JMI EEE 2001]

If tan $α$ = $\frac{m}{m + 1}$ and tan $β$ = $\frac{1}{2 m + 1}$ , then $α$ + $β$ =

[IIT 1978; EAMCET 1992; Roorkee 1998; JMI EEE 2001]