If tan -sin 2 -3-cos 2 - then

The correct option is A tanα=2 tanβ We have sin 2β/3 cos 2β=2 sinβ.cosβ/2 2cos2β+1+cos 2β =2 sinβ.cosβ/4 sin^2β+2 cos^2β=tanβ/1+2 tan^2β=2 tanβ tanβ/1+2 tan^2β ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

If tan α-β=si...

Question

If $t a n (α - β) = \frac{s i n 2 β}{3 - c o s 2 β}$ ,then

$t a n α = 2 t a n β$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$t a n β = 2 t a n α$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2 t a n α = 3 t a n β$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$3 t a n α = 2 t a n β$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$t a n α = 2 t a n β$

We have $\frac{s i n 2 β}{3 - c o s 2 β} = \frac{2 s i n β . c o s β}{2 - 2 c o s 2 β + 1 + c o s 2 β}$

$= \frac{2 s i n β . c o s β}{4 s i n^{2} β + 2 c o s^{2} β} = \frac{t a n β}{1 + 2 t a n^{2} β} = \frac{2 t a n β - t a n β}{1 + 2 t a n^{2} β}$

$= t a n (α - β) = \frac{t a n α - t a n β}{1 + t a n α . t a n β} = \frac{2 t a n β - t a n β}{1 + 2 t a n^{2} β}$

$∴ t a n α = 2 t a n β$

Suggest Corrections

Similar questions

Q. If

tan (α - β) = \frac{sin 2 β}{5 - cos 2 β}

, find the value of

\frac{tan α}{tan β}

If 2 $t a n α = 3 t a n β$ , prove that $t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$

Q. If

\frac{t a n (α + β - γ)}{t a n (α - β + γ)} = \frac{t a n γ}{t a n β}

, then either

s i n (β - γ) = 0

s i n 2 α + s i n 2 β + s i n 2 γ = 0

\frac{sin 2 α + sin 2 β}{cos 2 α - cos 2 β} = cot (β - α)

If $α a n d β$ are acute angles satisfying $c o s 2 α = \frac{3 c o s 2 β - 1}{3 - c o s 2 β},$ then $t a n α$ =

Join BYJU'S Learning Program

If tan(α−β)=sin 2β3−cos2β,then

The correct option is A tanα=2 tanβ We have sin 2β3−cos2β=2 sinβ.cosβ2−2cos2β+1+cos2β =2 sinβ.cosβ4 sin2β+2 cos2β=tanβ1+2 tan2β=2 tanβ−tanβ1+2 tan2β =tan(α−β)=tanα−tanβ1+tanα.tanβ=2 tanβ−tanβ1+2 tan2β ∴tanα=2 tanβ

If $t a n (α - β) = \frac{s i n 2 β}{3 - c o s 2 β}$ ,then