Sin 2 A -1-cos 2 A -A- tan AB- A2 tan A -2-1-tan 2 A -2D- tan 2 A

The correct options are A tan A C 2tanA/21 tan^2 A/2 nbsp; sin2A1 + cos2A Substituting the values of nbsp; sin 2A = 2sin A·cos A and cos 2A = 2cos^2 A 1 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

sin 2 A /1+co...

Question

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

$tan A$

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$cot A$

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$\frac{2 tan \frac{A}{2}}{1 - {tan}^{2} \frac{A}{2}}$

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$tan 2 A$

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Solution

The correct options are
A
$tan A$

C
$\frac{2 tan \frac{A}{2}}{1 - {tan}^{2} \frac{A}{2}}$

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

Substituting the values of $sin 2 A = 2 sin A \cdot cos A$ and $cos 2 A = 2 {cos}^{2} A - 1$

$= \frac{2 sin A . cos A}{1 + 2 {cos}^{2} A - 1}$

$= \frac{2 sin A . cos A}{2 {cos}^{2} A} = tan A$

We know, $tan 2 A = \frac{2 tan A}{1 - {tan}^{2} A}$

So, $tan A = \frac{2 tan \frac{A}{2}}{1 - {tan}^{2} \frac{A}{2}}$

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Similar questions

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

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

Q. Prove that:

\frac{{cos}^{2} A + {tan}^{2} A - 1}{{sin}^{2} A} = {tan}^{2} A .

Q. If A =

30^{\circ}

, then

cos 2 A = \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} = {cos}^{2} A - {sin}^{2} A

State true or false.

If $3 c o t A = 4 . Prove that \frac{1 - t a n^{2} A}{1 + t a n^{2} A} = c o s^{2} A - s i n^{2} A$ .

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sin2A1+cos2A =

The correct options are A tanA C 2tanA21−tan2A2 sin2A1+cos2A Substituting the values of sin2A=2sinA⋅cosA and cos2A=2cos2A−1 =2sinA.cosA1+2cos2A−1 =2sinA.cosA2cos2A=tanA We know, tan2A=2tanA1−tan2A So, tanA=2tanA21−tan2A2

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