If 2 tan - - 3 tan - prove that tan - - - - sin2 -5 - cos 2-

LHS #160; ⇒ tan(α β)=tanα tanβ1+tanα.tanβ=tanβ2+3tan^2β RHS #160; ⇒sin2β5 cos2β=2sinβ.cosβ6 2cos^2β Dividing numerator, denominator by ...

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Trigonometric Equations

If 2 tan α ...

Question

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

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2 \tan α = 3 \tan β,

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

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

\frac{tan α}{tan β}

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

2 tan β + cot β = tan α

cot β = 2 tan (α - β) .

2 \tan α = 3 \tan β, then \tan (α - β) =

\frac{\sin 2 β}{5 - \cos 2 β}

\frac{\cos 2 β}{5 - \cos 2 β}

\frac{\sin 2 β}{5 + \cos 2 β}

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