Is cot^2(θ)csc^2(θ)-cot^2(θ)=cot^4(θ) ?

The answer to whether cot^2(θ)csc^2(θ)-cot^2(θ)=cot^4(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cot^2(θ)csc^2(θ)-cot^2(θ)=cot^4(θ)

prove

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ) = cot^{4} (θ)

True

Solution steps

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ) = cot^{4} (θ)

Manipulating left side

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ)

Rewrite using trig identities

= cot^{4} (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (x - π) = - sin (x)

cos^{2} (x) + tan^{2} (x) cos^{2} (x) = 1

sec^{2} (x) - 1 = tan^{2} (x)

cos (x - \frac{π}{3}) + sin (\frac{π}{6} - x) = cos (x)

coth^{2} (x) - 1 = csch^{2} (x)

Is cot^2(θ)csc^2(θ)-cot^2(θ)=cot^4(θ) ?
The answer to whether cot^2(θ)csc^2(θ)-cot^2(θ)=cot^4(θ) is True