The correct option is A sin A sin3A1+cos2A=3sin A 4sin^3A1+2(1 2sin^2A)=sin A(3 4sin^2A)3 4sin^2A=sin A THerefore, Answer is sin A ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Sum of n Terms

sin 3A1+2cos ...

Question

$\frac{sin 3 A}{1 + 2 cos 2 A} =$

sin A

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

cos A

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

tan A

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

cot A

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

Open in App

Solution

The correct option is A $sin A$
$\frac{sin 3 A}{1 + cos 2 A} = \frac{3 sin A - 4 {sin}^{3} A}{1 + 2 (1 - 2 {sin}^{2} A)} = \frac{sin A (3 - 4 {sin}^{2} A)}{3 - 4 {sin}^{2} A} = sin A$

THerefore, Answer is $sin A$

Suggest Corrections

Similar questions

Q. Solve

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

Q. Prove the identity:

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{1 - 2 c o s^{2} A}

\frac{{cos}^{3} A + {sin}^{3} A}{cos A + sin A} + \frac{{cos}^{3} A - {sin}^{3} A}{cos A - sin A}

Q. Prove that :

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A} = \frac{2}{2 s i n^{2} A - 1} = \frac{2}{1 - 2 c o s^{2} A}

Join BYJU'S Learning Program

sin3A1+2cos2A=

The correct option is A sinAsin3A1+cos2A=3sinA−4sin3A1+2(1−2sin2A)=sinA(3−4sin2A)3−4sin2A=sinATHerefore, Answer is sinA

$\frac{sin 3 A}{1 + 2 cos 2 A} =$

The correct option is A $sin A$
$\frac{sin 3 A}{1 + cos 2 A} = \frac{3 sin A - 4 {sin}^{3} A}{1 + 2 (1 - 2 {sin}^{2} A)} = \frac{sin A (3 - 4 {sin}^{2} A)}{3 - 4 {sin}^{2} A} = sin A$

THerefore, Answer is $sin A$