Question

pls find answer. ........ prove that sec^6x-tan^6x=1+3tan^2x+×tan^2x​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

${sec}^{6} x - {tan}^{6} x = 1 + 3 {sec}^{2} x {sec}^{2} x \\ LHS = {sec}^{6} x - {tan}^{6} x \\ = ({sec}^{2} ) ^ {3} x - ({tan}^{2})^{3} x \\ = ({sec}^{2} x - {tan}^{2}x) ^{3} + 3 {sec}^{2} x \: {tan}^{2} x \: ({sec}^{2} x - {tan}^{2}x) \\ = (1) ^{3} + 3 {sec}^{2} x \:{tan}^{2} x (1) \\ = 1 + 3{sec}^{2}x \: {tan}^{2} x \\ = \: RHS \\Thus \: proved \\ \\ formula \: used : \\ {x}^{3} - {y}^{3} = (x - y) ^{3} + 3xy(x - y)$