Math, asked by Anonymous, 9 months ago

Prove that sec⁴A-sec²A=tan⁴A+tan²A(without spam)

Answers

Answered by yashaswini3679

Step-by-step explanation:

LHS

= sec⁴A-sec²A

= sec²A(sec²A - 1) ------- [sec²A - tan²A = 1]

= sec²A(tan²A)

= (tan²A + 1)(tan²A) ------ [sec²A - tan²A = 1]

= tan⁴A + tan²A

= RHS

Hence, proved.

HOPE THIS ANSWER HELPS U....

Answered by Anonymous

$\huge\boxed{\fcolorbox{white}{blue}{Prove that -}}$

sec⁴A-sec²A=tan⁴A+tan²A

______________________________

$\huge\boxed{\fcolorbox{blue}{pruple}{Given-}}$

sec⁴A - sec²A = L.H.S

tan⁴A + tan²A = R.H.S

______________________________

$\huge\boxed{\fcolorbox{yellow}{green}{Proof-}}$

L.H.S

= sec⁴A - sec² A

= (1 + tan²A)² - [ 1+tan²A ]

= 1 + 2 tan²A+ tan²A - 1 - tan²A

= tan⁴A + tan²A

= R.H.S

_______________________________

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