Prove that- cos2A - tan2A - 1sin2A - tan 2A-

LHS =cos^2A+tan^2A 1sin^2A cos^2A cosec^2A+^2A =1+^2A cosec^2A 1+^2A =cosec^2A cosec^2A+^2A 1 [1+^2A=cosec^2A] =^2A 1 [1+tan^2A ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Basic Trigonometric Identities

Prove that: ...

Question

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

Open in App

Solution

Suggest Corrections

Similar questions

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

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

Join BYJU'S Learning Program