Is tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ) ?

The answer to whether tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ) is True

Popular >

prove tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ)

prove

tan^{2} (θ) - sin^{2} (θ) = sin^{2} (θ) tan^{2} (θ)

True

Solution steps

tan^{2} (θ) - sin^{2} (θ) = sin^{2} (θ) tan^{2} (θ)

Manipulating right side

sin^{2} (θ) tan^{2} (θ)

Rewrite using trig identities

= (1 - cos^{2} (θ)) tan^{2} (θ)

Expand

(1 - cos^{2} (θ)) tan^{2} (θ) : tan^{2} (θ) - tan^{2} (θ) cos^{2} (θ)

= tan^{2} (θ) - cos^{2} (θ) tan^{2} (θ)

Rewrite using trig identities

= tan^{2} (θ) - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (2 u) - sin^{2} (2 u) = cos (4 u)

tan (u) cot (u) - cos^{2} (u) = sin^{2} (u)

\frac{1 - sin ^{2} ( x )}{csc ^{2} ( x ) - 1} = sin^{2} (x)

cos (2 θ) = 1 - 2 sin^{2} (θ)

sin (x) = 1 - cos (x)

Is tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ) ?
The answer to whether tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ) is True