Get the answer to "tan2⁡θ−sin2⁡θ=tan2⁡θsin⁡θ"

Emeli Hagan

Answered question

2021-09-09

$\tan^{2} θ - \sin^{2} θ = \tan^{2} θ \sin θ$

okomgcae

Skilled2021-09-10Added 93 answers

Work on the left side. Use the quotient identity for tangent: $\tan^{2} θ - \sin^{2} θ = (\frac{\sin^{2} θ}{\cos^{2} θ}) - \sin^{2} θ$

Factor out $\sin 2 θ$ : $\tan^{2} θ - \sin^{2} θ = \sin^{2} θ ((\frac{1}{\cos^{2} θ}) - 1)$

$\tan^{2} θ - \sin^{2} θ = \sin^{2} θ (\frac{1 - \cos^{2} θ}{\cos^{2} θ})$

Use the Pythagorean identity: $\sin^{2} θ + \cos^{2} θ = 1$

$\tan^{2} θ - \sin^{2} θ = \sin^{2} θ (\frac{\sin^{2} θ}{\cos^{2} θ})$

$\tan^{2} θ - \sin^{2} θ = \sin^{2} θ \tan^{2} θ$

or $\tan^{2} θ - \sin^{2} θ = \tan^{2} θ \sin^{2} θ$

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?