Question

Prove the following trigonometric identity: (1+sin^(4)A)/(cos^(4)A)=1+2tan^(2)A*sec^(2)A

Answer 1

GIVEN:

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

TO PROVE:

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

FORMULAE:

Sin² A = 1 - Cos² A

Tan A = Sin A / Cos A

Sec A = 1 / Cos A

(A - B)² = A² - 2AB + B²

PROOF:

$\frac{1 + \sin^4 A }{ { \cos}^{4} A} = 1 + 2( { \tan}^{2} A \times { \sec}^{2}A) \\ \\ \frac{1 + {(1 - { \cos}^{2} A )}^{2} }{ { \cos}^{4} A } = 1 + 2( \frac{ { \sin}^{2} A }{ { \cos}^{2} A } \times \frac{1}{ { \cos}^{2} A } ) \\ \\ \frac{{1 + 1 + { \cos}^{4} A - 2 { \cos}^{2} A }}{ { \cos}^{4} A } = 1 + 2( \frac{ { \sin}^{2} A }{ { \cos}^{4} A }) \\ \\\frac{{2 + { \cos}^{4} A - 2 { \cos}^{2} A }}{ { \cos}^{4} A } = \frac{ { \cos}^{4} A + 2 { \sin}^{2} A }{ { \cos}^{4} A } \\ \\ 2 + { \cos}^{4} A - 2 { \cos}^{2} A = { \cos}^{4} A + {2 \sin}^{2} A \\ \\ 2 - 2 { \cos}^{2} A = 2{ \sin}^{2} A \\ \\ 1 - { \cos}^{2} A = { \sin}^{2} A \\ \\ L.H.S=R.H.S$

HENCE PROVED.