Show that tan²θ+ tan⁴θ = sec⁴θ − sec²θ
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Answered by
10
L.H.S. = tan²θ + tan⁴θ
= tan²θ (1 + tan²θ)
= tan²θ . sec²θ
= (sec²θ − 1) sec²θ
= sec⁴θ − sec²θ
= R.H.S
hence proved
= tan²θ (1 + tan²θ)
= tan²θ . sec²θ
= (sec²θ − 1) sec²θ
= sec⁴θ − sec²θ
= R.H.S
hence proved
SparshGupta:
this is not sec²@ - sec²@. please correct your answer
Answered by
3
tan²@ + tan⁴@ = sec⁴@ - sec²@
tan²@ ( 1 + tan²@ ) = RHS
( sec²@ - 1 )( sec²@ ) = RHS
sec⁴@ - sec²@ = RHS
LHS = RHS
hope this help you......
Asking questions is a sign of intelligency.
tan²@ ( 1 + tan²@ ) = RHS
( sec²@ - 1 )( sec²@ ) = RHS
sec⁴@ - sec²@ = RHS
LHS = RHS
hope this help you......
Asking questions is a sign of intelligency.
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