Prove 2 Trig Identities (1 Viewer)

[Utility]

Prove LHS equals the RHS:

1. (sec²Ѳ - 1)cos²Ѳ = sin²Ѳ

2. 1 / cot²Ѳ + 1 = 1 / cos²Ѳ

 

[Shadowdude]

1. Trick is: sec(x) = 1/cos(x)
2. Trick is: cot(x) = 1/tan(x)

With those, you should be able to do the rest. Have fun!

 

[OH1995]

1. (sec^2ø-1)cos^2ø=sin^2ø
tan^2øcos^2ø=sin^2ø (using identity)
tan^2ø=sin^2ø/cos^2ø (dividing by cos)
therefore tan^2ø=tan^2ø

 

[theind1996]

Prove LHS equals the RHS:

1. (sec²Ѳ - 1)cos²Ѳ = sin²Ѳ

2. 1 / cot²Ѳ + 1 = 1 / cos²Ѳ

1. (sec²Ѳ - 1)cos²Ѳ = sin²Ѳ

LHS = tan²Ѳcos²Ѳ

= (sin²Ѳ/cos²Ѳ)cos²Ѳ

Cosines cancel each other

= sin²Ѳ
= RHS

2. 1 / cot²Ѳ + 1 = 1 / cos²Ѳ

LHS = tan²Ѳ+1

= sec²Ѳ

= 1/cos²Ѳ

= RHS

 

[OH1995]

(1/cot^2ø)+1=(1/cos^2ø)
tan^2ø+1=sec^2ø (reciprocal functions)
Therefore, sec^2ø=sec^2ø (trig identity: tan^2ø+1=sec^2ø)