What is $\frac{{sin}^{2} θ}{1 - tan θ}$ $sin^2theta/(1-tantheta)$ in terms of $cos θ$ $costheta$ ?

Question

1625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-03T05:43:57

$cos θ \frac{(1 - {cos}^{2} θ)}{cos θ - \sqrt{1 - {cos}^{2} θ}}$ $costheta((1−cos^2theta))/(costheta-sqrt(1−cos^2theta))$

$\frac{{sin}^{2} θ}{1 - tan θ}$ $sin^2theta/(1−tantheta)$ can be written in term of $cos θ$ $costheta$ , using identities

${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2theta=1−cos^2theta$ i.e. $sin θ = \sqrt{1 - {cos}^{2} θ}$ $sintheta=sqrt(1−cos^2theta)$

and $tan θ = \frac{sin θ}{cos θ}$ $tantheta=sintheta/costheta$

As such $\frac{{sin}^{2} θ}{1 - tan θ}$ $sin^2theta/(1−tantheta)$

= $\frac{1 - {cos}^{2} θ}{1 - \frac{sin θ}{cos θ}}$ $(1−cos^2theta)/(1-sintheta/costheta)$

Multi[lying numerator and denominator by $cos θ$ $costheta$ , we get

$cos θ \frac{(1 - {cos}^{2} θ)}{cos θ - sin θ}$ $costheta((1−cos^2theta))/(costheta-sintheta)$ or

$cos θ \frac{(1 - {cos}^{2} θ)}{cos θ - \sqrt{1 - {cos}^{2} θ}}$ $costheta((1−cos^2theta))/(costheta-sqrt(1−cos^2theta))$

What is sin^2theta/(1-tantheta) sin2θ1−tanθsin^2theta/(1-tantheta) in terms of costhetacosθcostheta?