Asked by Alycia
How do I do these questions? I'm supposed to prove them so that both sides are the same and I'm only supposed to manipulate one side and leave the other side alone but I'm stuck.
Problem 1. (secx/cosx)-(cscx/sinx)=tan^2x-cot^2x
Problem 2. secx+cscx=(1+tanx)/sinx
Problem 1. (secx/cosx)-(cscx/sinx)=tan^2x-cot^2x
Problem 2. secx+cscx=(1+tanx)/sinx
Answers
Answered by
Reiny
1.
LS = (1/cosx) / cosx - (1/sinx) /sinx
= 1/cos^2 x - 1/(sin^2 x)
= (sin^2x + cos^2x)/cos^2x - (sin^2x + cos^2x)/sin^2x
= sin^2x/cos^2x + cos^2x/cos^2x - sin^2x/sin^2x - cos^2x/sin^2x
= tan^2x + 1 - 1 - cot^2x
= tan^2x - cot^2x
= RS
2. RS = (1+tanx)/sinx
= 1/sinx + tanx/sinx
= cscx + (sinx/cosx) /sinx
= cscx + 1/cosx
= cscx + secx
= LS
LS = (1/cosx) / cosx - (1/sinx) /sinx
= 1/cos^2 x - 1/(sin^2 x)
= (sin^2x + cos^2x)/cos^2x - (sin^2x + cos^2x)/sin^2x
= sin^2x/cos^2x + cos^2x/cos^2x - sin^2x/sin^2x - cos^2x/sin^2x
= tan^2x + 1 - 1 - cot^2x
= tan^2x - cot^2x
= RS
2. RS = (1+tanx)/sinx
= 1/sinx + tanx/sinx
= cscx + (sinx/cosx) /sinx
= cscx + 1/cosx
= cscx + secx
= LS
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