Question

1-sin^4θ/cos^4θ=2sec^2 θ -1 Prove by taking LHS

Answer 1

hope I had solved u r problem

Answer 2

AnswEr :

• To Prove :

$\sf \dfrac{1 - { \sin}^{4} ( \theta)}{ { \cos}^{4} ( \theta)} = 2 { \sec }^{2}( \theta) - 1$

• Proof :

$\leadsto \rm \dfrac{1 - { \sin}^{4} ( \theta)}{ { \cos}^{4} ( \theta)}$

$\leadsto \rm \dfrac{(1)^{2} - ({ \sin}^{2}( \theta))^{2} }{ { \cos}^{4} ( \theta)}$

⠀⠀⠀⠀⋆ (a² - b²) = (a + b)(a - b)

$\leadsto \rm \dfrac{(1 + { \sin}^{2} ( \theta))(1 - { \sin}^{2} ( \theta))}{ { \cos}^{4} ( \theta)}$

⠀⠀⠀⠀⋆ (1 - sin² θ = cos² θ)

$\leadsto \rm \dfrac{(1 + { \sin}^{2} ( \theta)) \cancel{{\cos}^{2} ( \theta)}}{ \cancel{{ \cos}^{4} ( \theta)}}$

$\leadsto \rm \dfrac{(1 + { \sin}^{2} ( \theta))}{{ \cos}^{2} ( \theta)}$

$\leadsto \rm\dfrac{1}{{ \cos}^{2} ( \theta)} + \dfrac{{ \sin}^{2} ( \theta)}{{ \cos}^{2} ( \theta)}$

⠀⠀⠀⠀⋆ 1 / cos² θ = sec² θ

⠀⠀⠀⠀⋆ sin² θ / cos² θ = tan² θ

$\leadsto \rm \sec^{2} ( \theta) + \tan^{2} ( \theta)$

⠀⠀⠀⠀⋆ tan² θ = (sec² θ - 1)

$\leadsto \rm \sec^{2} ( \theta) + \sec^{2} ( \theta) - 1$

$\leadsto \large\rm 2\sec^{2} ( \theta) - 1$

⠀

$\therefore \boxed{ \sf \dfrac{1 - { \sin}^{4} ( \theta)}{ { \cos}^{4} ( \theta)} = 2 { \sec }^{2}( \theta) - 1}$