Question

sin⁴A+sin²A=1 if than prove th
at tan⁴A- tan²A =1​

Answer 1

Step-by-step explanation:

${sin}^{4} x + {sin}^{2} x = 1$

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

HENCE PROVED.

Answer 2

Answer:

Step-by-step explanation:

sin⁴A+sin²A =1

sin⁴A=1-sin²A

tan⁴A=cos²A

Taking,

L.H.S,

tan⁴A-tan²A

sin⁴A/cos⁴A - sin²A/cos²A

1/cos²A- sin²A/cos²A

1-sin²A/cos²A

cos²A/cos²A

1