Question

cot^2 θ − cos^2 θ = cos^2 θ cot^2 θ proof please help >~<

Answer 1

Is it \(cot^2 \theta\) or \(cot2 \theta\)??

Answer 2

Yes got it..

Answer 3

how do you solve it? I'm stuck

Answer 4

cot^2 θ − cos^2 θ=cos^2 θ((1/sin^2 θ) -1)
=cos^2 θ(cosec^2 θ-1)=cos^2 θcot^2 θ

Answer 5

replace \(a\) by \(\cos(x)\) and \(b\) by \(sin(x)\) and do some algebra

\[\frac{a^2}{b^2}-a^2\]
\[=\frac{a^2-a^2b^2}{b^2}\]
\[=\frac{a^2(1-b^2)}{b^2}\] replace \(a\) by \(\cos(x)\) and \(b\) by \(sin(x)\) and do some algebra

Answer 6

what about the other side? if you started off on with the cos^2θ cot^2θ?

I'm doing it online and its fill in the the blank kind of deal. The part that I'm at looks like this
cos^2θ cot^2θ =(________) (cot^2θ)
= cot^2 θ −(________)(cot^2θ)
Use a Reciprocal Identity, and then simplify.
= cot^2 θ − (________)cos^2θ/sin^2θ
= ___________

Answer 7

then a little trig. since \(b=\sin(x)\), then \(1-b^2=1-\sin^2(x)=\cos^2(x)\)
giving
\[\frac{\cos^2(x)\cos^2(x)}{\sin^2(x)}=\cos^2(x)\cot^2(x)\]

Answer 8

cos^2θ cot^2θ =(__cos^2θ______) (cot^2θ)
= cot^2 θ −(___sin^2θ_____)(cot^2θ) Use a Reciprocal Identity, and then simplify. = cot^2 θ − (__sin^2θ______)cos^2θ/sin^2θ = _cot^2 θ − cos^2 θ __________

Answer 9

Using these identities:

\[\cot \theta = \frac{1}{\tan \theta} = \frac{cos \theta}{sin \theta}\]

\[\cos^2 \theta = 1 - \sin^2 \theta\]

\[\frac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta = \frac{(\cos^2 \theta)(1- \sin^2 \theta)}{\sin^2 \theta} = \cot^2 \theta \times \cos^2 \theta\]

Answer 10

cos^2θ cot^2θ =(1-sin^2θ) (cot^2θ) = cot^2 θ −(sin^2θ)(cot^2θ) Use a Reciprocal Identity, and then simplify. = cot^2 θ − (sin^2θ)cos^2θ/sin^2θ = cot^2 θ − cos^2 θ

Answer 11

Yes finally! thanks you all save me alot of time

Answer 12

mention not