Question

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $si{n}^{2}A-co{s}^{2}A$

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

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

5. $ta{n}^{2}A$ = $\frac{1-cos2A}{1+cos2A}$

6. $si{n}^{2}A$ = $\frac{2tanA}{1-ta{n}^{2}A}$

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Answer 1

Let's check for each statement.

Given sin2A = 2 sinA. cosA

We know sin(A+B) = sinA.cosB + cosA.sinA

If B = A

Sin (A+A) = sinA . cosA + cosA . sinA

sin2A = 2sinA cosA

statement 1 is correct.

cos2A = $si{n}^{2}A-co{s}^{2}A$

We know cos(A+B) = cosA.cosB - sinA.sinB

If B = A, then we get A + A = 2A

cos (A+A) = cosA . cosA - sinA . sinA

sin2A = $co{s}^{2}A-si{n}^{2}A$

statement 2 is incorrect.

Statement : 3 tanA = $\frac{2tan\frac{A}{2}}{1-ta{n}^2\frac{A}{2}}$

We can derive this using the formula for tan(A+B). In that,replace A with B to get tan2A.We will find statement 3 is correct.

Statement 4: $si{n}^2\frac{A}{2}$ = 1 - cosA

We know,cos 2A = $co{s}^{2}A-si{n}^{2}A$ = 1 - $si{n}^{2}A-si{n}^{2}A$ = 1 - $2si{n}^{2}A$

Similarly, cos A = 1 - 2$si{n}^2\frac{A}{2}$

$2si{n}^2\frac{A}{2}$ = 1 - cosA

$si{n}^2\frac{A}{2}$ = $\frac{1-cosA}{2}$

So,statement 4 is in correct

Statement 5 : $ta{n}^{2}A$ = $\frac{1-cos2A}{1+cos2A}$

Statement contains cos2A terms. So,Let's check for cos2A.

We know, cos2A = $co{s}^{2}A-si{n}^{2}A$ = 1 - $2si{n}^{2}A=2co{s}^{2}A-1$

$\Rightarrow$ $\frac{1-cos2A}{1+Cos2A}$ = $\frac{2si{n}^{2}A}{2co{s}^{2}A}$

= $ta{n}^{2}A$.

Statement 6: sin2A = $\frac{2tanA}{1-ta{n}^{2}A}$ gives tanA in statement 3

Only statement 1,3 & 5 are correct.