Question

Prove the following______​

Answer 1

Short Answer

1+tan²x=sec²x

(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)

                          =1³+3sec²xtan2x(1)

                            =1+3tan²xsex²x

Long Answer

Recall that 1 + tan2x = sec2x  and A3-B3 = (A-B)(A2+AB+B2)

sec6x - tan6x = (sec2x)3 - (tan2x)3

            $= (sec2x - tan2x)(sec4x + sec2xtan2x + tan4x)$

                  = (1+tan2x-tan2x))[sec4x+sec2xtan2x+(sec2x-1)tan2x]

                  = sec4x - tan2x + 2sec2xtan2x

                  = sec2x(1+tan2x) - tan2x + 2sec2xtan2x

                  = sec2x - tan2x + sec2xtan2x + 2sec2xtan2x

                  = 1 + 3sec2xtan2x

Hope it helps

KL mumbai jit gyi

#Jasprit bumrah

#Trent boult

Answer 2

Step-by-step explanation:

sec6x-tan6x=(sec²x)³−(tan²x)³

={sec²x−tan²x}³ +3sec²x.tan²x(sec² x−tan²x)

=1+3sec²x.tan²x.

[ Since sec²x−tan²x=1]

i hope this help u