If 2 tan-3 tan- then tan -

The correct option is A sin 2 β/5 cos 2 β 2 tan α = 3 tan β Now, tan(α β)=tanα tan β/1+tan α tan β =3/2 tan β tan β/1+(3/2tan β )tan β =3 tan β 2 tan β/2+3 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If 2 tanα=3 t...

Question

If 2 tan $α$ = 3 tan $β$ , then tan $(α - β)$ =

$\frac{s i n 2 β}{5 - c o s 2 β}$

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$\frac{c o s 2 β}{5 - c o s 2 β}$

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$\frac{s i n 2 β}{5 + c o s 2 β}$

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none of these

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Solution

The correct option is A
$\frac{s i n 2 β}{5 - c o s 2 β}$

$2 t a n α = 3 t a n β$

Now,

$t a n (α - β) = \frac{t a n α - t a n β}{1 + t a n α t a n β} = \frac{\frac{3}{2} t a n β - t a n β}{1 + (\frac{3}{2} t a n β) t a n β} = \frac{3 t a n β - 2 t a n β}{2 + 3 t a n^{2} β} = \frac{t a n β}{2 + 3 t a n^{3} β} = \frac{\frac{s s i n β}{c o s β}}{2 + 3 \frac{s i n^{2} β}{c o s^{2} β}}$

$= \frac{s i n β c o s β}{2 c o s^{2} β + 3 s i n^{2} β} = \frac{s i n β c o s β}{3 + s i n^{2} β} = \frac{2 s i n β c o s β}{4 + 2 + s i n^{2} β}$

$= \frac{s i n 2 β}{4 + 1 - c o s 2 β} ∴ t a n (α - β) = \frac{s i n 2 β}{5 - c o s 2 β}$

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Similar questions

Q. If

2 \tan α = 3 \tan β,

prove that

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

Q. If tan

α = 3 tan β

, then the maximum value

{tan}^{2} (α - β)

Q. 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

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

(a)

\frac{\sin 2 β}{5 - \cos 2 β}

(b)

\frac{\cos 2 β}{5 - \cos 2 β}

(c)

\frac{\sin 2 β}{5 + \cos 2 β}

(d) none of these

Q. If

2 tan β + cot β = tan α

, then prove that

cot β = 2 tan (α - β)

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If 2 tan α = 3 tan β, then tan (α−β)=

If 2 tan $α$ = 3 tan $β$ , then tan $(α - β)$ =