How do you express $cos θ - {csc}^{2} θ + sec θ$ $cos theta - csc^2 theta + sec theta$ in terms of $sin θ$ $sin theta$ ?

Question

How do you express $cos θ - {csc}^{2} θ + sec θ$ $cos theta - csc^2 theta + sec theta$ in terms of $sin θ$ $sin theta$ ?

1 Answer

Impact of this question

1590 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-02T15:01:14

$\frac{2 {sin}^{2} θ - {sin}^{4} θ - \sqrt{1 - {sin}^{2} θ}}{{sin}^{2} θ \sqrt{1 - {sin}^{2} θ}}$ $(2sin^2theta -sin^4theta -sqrt(1-sin^2theta))/(sin^2thetasqrt(1-sin^2theta))$

Explanation:

Using trigonometric relationships we can say

$cos θ - {csc}^{2} θ + sec θ$ $costheta - csc^2theta+sectheta$
$= cos θ - \frac{1}{{sin}^{2} θ} + \frac{1}{cos θ}$ $=costheta -1/sin^2theta +1/costheta$
$= \frac{{cos}^{2} θ + 1}{cos θ} - \frac{1}{{sin}^{2} θ}$ $=(cos^2theta+1)/costheta - 1/sin^2theta$

We know that ${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta =1-sin^2theta$

Substituting this into tthe expression gives
$\frac{1 - {sin}^{2} θ + 1}{cos θ} - \frac{1}{{sin}^{2} θ}$ $(1-sin^2theta+1)/costheta -1/sin^2theta$

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

$cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta = sqrt(1 - sin^2theta)$

Substituting again gives

$\frac{2 {sin}^{2} θ - {sin}^{4} θ - \sqrt{1 - {sin}^{2} θ}}{{sin}^{2} θ \sqrt{1 - {sin}^{2} θ}}$ $(2sin^2theta -sin^4theta -sqrt(1-sin^2theta))/(sin^2thetasqrt(1-sin^2theta))$

Science

Math

Humanities

... and beyond

How do you express $cos θ - {csc}^{2} θ + sec θ$ $cos theta - csc^2 theta + sec theta$ in terms of $sin θ$ $sin theta$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you express cosθ−csc2θ+secθcos theta - csc^2 theta + sec theta in terms of sinθsin theta ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you express $cos θ - {csc}^{2} θ + sec θ$ $cos theta - csc^2 theta + sec theta$ in terms of $sin θ$ $sin theta$ ?