Tan 3A-tan A-sin 3A-sin A-

The correct option is B 2a/a 1 tan 3A #160; tan A #160; =a 2a a 1 = 2tan 3A #160; tan A #160; #160; tan 3A #160; tan A # ...

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Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

tan 3A/tan A⇒...

Question

$\frac{t a n 3 A}{t a n A} \Rightarrow \frac{s i n 3 A}{s i n A} =$

\frac{2 a}{a + 1}

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\frac{2 a}{a - 1}

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\frac{a}{a + 1}

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\frac{a}{a - 1}

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Solution

The correct option is B $\frac{2 a}{a - 1}$
$\frac{tan 3 A}{tan A} = a$
$\frac{2 a}{a - 1} = \frac{\frac{2 tan 3 A}{tan A}}{\frac{tan 3 A}{tan A} - 1} = \frac{2 tan 3 A}{tan 3 A - tan A}$
$= \frac{2 sin 3 A cos A}{cos A sin 3 A - sin A cos 3 A} = \frac{2 sin 3 A cos A}{sin 2 A}$
$= \frac{2 sin 3 A cos A}{2 sin A cos A} = \frac{sin 3 A}{sin A}$

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Similar questions

$\frac{1}{t a n 3 A - t a n A} - \frac{1}{c o t 3 A - c o t A} =$

Q. Prove the following :

\frac{1}{tan 3 A - tan A} - \frac{1}{cot 3 A - cot A} = cot 2 A

Q. If

\frac{tan 3 A}{tan A} = a,

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\frac{sin 3 A}{sin A} =

Q. Prove that :

\frac{{tan}^{3} A - {cot}^{3} A}{tan A - cot A} = {sec}^{2} A + {cot}^{2} A .

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

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tan3AtanA⇒sin3AsinA=

The correct option is B 2aa−1tan3AtanA=a2aa−1=2tan3AtanAtan3AtanA−1=2tan3Atan3A−tanA=2sin3AcosAcosAsin3A−sinAcos3A=2sin3AcosAsin2A=2sin3AcosA2sinAcosA=sin3AsinA

$\frac{t a n 3 A}{t a n A} \Rightarrow \frac{s i n 3 A}{s i n A} =$