Prove that- cos 2 A-tan 2 A-1 sin 2 A -tan 2 A-

LHS=cos^ 2 A+tan^ 2 A 1 sin^ 2 A =tan ^2 A (1 cos ^2 A)sin ^2 A=tan ^2 A sin ^2 Asin ^2 A=sin ^2 Acos^2 Asin^2 A 1=sec^2 A 1=tan^ 2 A=RHS ...

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Use of Trigonometric Table

Prove that: ...

Question

Prove that: $\frac{{cos}^{2} A + {tan}^{2} A - 1}{{sin}^{2} A} = {tan}^{2} A$ .

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t a n^{2} A - t a n^{2} B = \frac{c o s^{2} B - c o s^{2} A}{c o s^{2} B c o s^{2} A} = \frac{s i n^{2} A - s i n^{2} B}{c o s^{2} A c o s^{2} B}

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

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

30^{\circ}

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

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

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