Question

prove the following​

Answer 1

Step-by-step explanation:

Solution is attached as a pic for ur reference.

Answer 2

Answer:-

To Prove:

$\sf \large \: \sec ^{2} A - \frac{ { \sin}^{2}A - 2 { \sin}^{4} A}{2 \cos ^{4} A - { \cos}^{2} A} = 1$

$\implies \sf \: { \sec}^{2}A - \dfrac{ { \sin }^{2} A(1 - 2 \sin ^{2}A) }{ { \cos}^{2} A(2 \cos ^{2}A - 1)} = 1$

We know that,

sin² A + cos² A = 1

⟹ sin² A = 1 - cos² A

$\implies \sf \: { \sec}^{2} A- \dfrac{ { \sin }^{2}A \{1 - 2(1 - { \cos \: }^{2} A)\}}{ { \cos }^{2} (2 { \cos}^{2}A - 1) } = 1 \\ \\ \implies \sf \: { \sec}^{2} A - \frac{ { \sin }^{2} A(1 - 2 + 2 { \cos }^{2} A - 1)}{ { \cos}^{2} A (2 { \cos }^{2} A - 1) } = 1 \\ \\ \implies \sf \: { \sec}^{2} A - \frac{ { \sin }^{2} A \cancel{ (2cos ^{2} A - 1)}}{ { \cos}^{2} A \cancel{(2cos ^{2} A - 1)} } = 1$

using sin² A/cos² = tan² A we get,

$\implies \sf \: { \sec}^{2} A - \tan ^{2} A = 1$

using the identity sec² A - tan² A = 1 in LHS we get,

⟹ 1 = 1.

Hence, Proved.