Question

(1 + sin^2 A )/cos^2 A= sec^4 A - tan^4 A​

Answer 1

Solution :-

$\frac{1 + {sin}^{2} a}{ {cos}^{2} a} = {sec}^{4} a - {tan}^{4} a$

Let's take L.H.S. →

$= \frac{1 + {sin}^{2} a}{ {cos}^{2} a}$

Divide multiply by (1 - sin²a) →

$= \frac{(1 + {sin}^{2}a)(1 - {sin}^{2} a)}{ {cos}^{2}a(1 - {sin}^{2} a) } \\ \\ = \frac{1 - {sin}^{4}a }{ {cos}^{2}a \times {cos}^{2} a } \\ \\ = \frac{1 - {sin}^{4}a }{ {cos}^{4}a } \\ \\ = \frac{1}{ {cos}^{4} a} - \frac{ {sin}^{4} a}{ {cos}^{4} a} \\ \\ = {sec}^{4} a - {tan}^{4} a$

R.H.S

Hence proved !!

Identity used :-

• ${a}^{2} - {b}^{2} = (a + b)(a - b)$

• $1 - {sin}^{2} a = {cos}^{2} a$

• $\frac{1}{cos \: a} = sec \: a$

• $\frac{sin \: a}{cos \: a} = tan \: a$