Sin 2 A -1-cos 2 A -A- sin AB- AC- 2 tan A -2-1-tan 2 A -2D- tan 2 A

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Trigonometric Ratios of Multiples of an Angle

sin 2 A /1+co...

Question

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

sin 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 option is C $\frac{2 tan \frac{A}{2}}{1 - {tan}^{2} \frac{A}{2}}$
Substituting the values of $sin 2 A = 2 s i n A cos A$ and $cos 2 A = 2 {cos}^{2} A - 1$ , we get

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

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

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

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

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

If A = $30^{\circ},$ verify that:

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

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.

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sin2A1+cos2A=

The correct option is C 2tanA21−tan2A2 Substituting the values of sin2A=2sinAcosA and cos2A=2cos2A−1, we get 2sinAcosA1+2cos2A−1 =2sinAcosA2cos2A=tanA We know, tan2A=2tanA1−tan2A So, tanA=2tanA21−tan2A2

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