If tan- -3tan- - then- - -

The correct option is C 2sin2β1+2cos2β Consider given the function #10; #10;If tanβ =3tanα ,then(α +β ) = #10; #10;We know that #10; #10; ...

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Standard XII

Mathematics

Standard Formulae - 3

If tanβ =3t...

Question

If $tan β = 3 tan α, t h e n (α + β) =$

\frac{2 sin β}{1 - 2 cos β}

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\frac{2 sin 2 β}{1 + 2 cos 2 β}

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\frac{2 sin 2 β}{1 - 2 cos 2 β}

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None of these

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Solution

The correct option is C $\frac{2 sin 2 β}{1 + 2 cos 2 β}$
Consider given the function

If $tan β = 3 tan α, t h e n (α + β)$ =

We know that

$tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}$

$= \frac{\frac{tan β}{3} + tan β}{1 - \frac{tan β}{3} tan β}$

$= \frac{\frac{4 tan β}{3}}{\frac{3 - {tan}^{2} β}{3}}$

$= \frac{4 tan β}{3 - {tan}^{2} β}$

$= \frac{4 \frac{sin β}{cos β}}{3 - \frac{{sin}^{2} β}{{cos}^{2} β}}$

$= \frac{4 \frac{sin β}{cos β}}{\frac{3 {cos}^{2} β - {sin}^{2} β}{{cos}^{2} β}}$

$= \frac{4 sin β cos β}{3 {cos}^{2} β - (1 - {cos}^{2} β)}$

$= \frac{2 sin 2 β}{3 {cos}^{2} β - 1 + {cos}^{2} β}$

$= \frac{2 sin 2 β}{4 {cos}^{2} β - 1 - 2 + 2}$

$= \frac{2 sin 2 β}{2 (2 {cos}^{2} β - 1) + 1}$

$= \frac{2 sin 2 β}{2 cos 2 β + 1}$

$= \frac{2 sin 2 β}{1 + 2 cos 2 β}$
Hence, this is the answer.

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If tanβ=3tanα, then(α+β)=

If $tan β = 3 tan α, t h e n (α + β) =$