Asked by CARL
Prove that:
(tan²Θ)x(sin²Θ) = (tan²Θ)-(sin²Θ)
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(tan²Θ)x(sin²Θ) = (tan²Θ)-(sin²Θ)
show work
Answers
Answered by
Reiny
LS = (sin^2 Ø)/cos^2 Ø) * sin^2 Ø
= sin^4 Ø/cos^2 Ø
RS = sin^2 Ø / cos^2 Ø - sin^2 Ø
= (sin^2 Ø - sin^2 Ø(cos^2 Ø) )/cos^2 Ø
= sin^2 Ø( 1 - cos^2 Ø)/cos^2 Ø
= sin^2 Ø (sin^2 Ø) /cos^ Ø
= sin^4 Ø / cos^2 Ø
= LS
= sin^4 Ø/cos^2 Ø
RS = sin^2 Ø / cos^2 Ø - sin^2 Ø
= (sin^2 Ø - sin^2 Ø(cos^2 Ø) )/cos^2 Ø
= sin^2 Ø( 1 - cos^2 Ø)/cos^2 Ø
= sin^2 Ø (sin^2 Ø) /cos^ Ø
= sin^4 Ø / cos^2 Ø
= LS
Answered by
CARL
Thank You!!!
Answered by
Scotto
True/False.
sec Θ ° > tan Θ ° for every angle Θ.
Please explain why?
sec Θ ° > tan Θ ° for every angle Θ.
Please explain why?