Asked by melissa
prove these identies
sin^2+tan^2=sec^2-cos^2
sin^2 sec^2 +sin^2=tan^2+sin^2
sin^2+tan^2=sec^2-cos^2
sin^2 sec^2 +sin^2=tan^2+sin^2
Answers
Answered by
Reiny
Your equations contain only trig operators but they have no arguments
e.g. sin^2 is meaningless, it has to be something like sin^2Ø or sin^2x
Your expression is as meaningless as trying to evaluate
5 + √
anyway .....
for sin^2+tan^2=sec^2-cos^2
LS = sin^2x + sin^2x/cos^2x
= (sin^2x cos^2x + sin^2x)/cos^2x
= sin^2x(cos^2x + 1)/cos^2x
= (1 - cos^2x)(1+cos^2x)/cos^2x
= (1 - cos^4x)/cos^2x
RS = 1/cos^2x - cos^2x
= (1 - cos^4 x)/cos^2x
= LS
the 2nd:
LS = sin^2x(1/cos^2x) + sin^2x
= tan^2x + sin^2x
= RS
e.g. sin^2 is meaningless, it has to be something like sin^2Ø or sin^2x
Your expression is as meaningless as trying to evaluate
5 + √
anyway .....
for sin^2+tan^2=sec^2-cos^2
LS = sin^2x + sin^2x/cos^2x
= (sin^2x cos^2x + sin^2x)/cos^2x
= sin^2x(cos^2x + 1)/cos^2x
= (1 - cos^2x)(1+cos^2x)/cos^2x
= (1 - cos^4x)/cos^2x
RS = 1/cos^2x - cos^2x
= (1 - cos^4 x)/cos^2x
= LS
the 2nd:
LS = sin^2x(1/cos^2x) + sin^2x
= tan^2x + sin^2x
= RS
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