How do you verify the identify $sin θ tan θ + cos θ = sec θ$ $sinthetatantheta+costheta=sectheta$ ?

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How do you verify the identify $sin θ tan θ + cos θ = sec θ$ $sinthetatantheta+costheta=sectheta$ ?

3 Answers

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Answer 1 · 2018-03-09T14:02:04

See the proof below

Explanation:

We need

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

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

$sec θ = \frac{1}{cos θ}$ $sectheta=1/costheta$

Therefore,

$L H S = sin θ tan θ + cos θ$ $LHS=sinthetatantheta+costheta$

$= sin θ \cdot \frac{sin θ}{cos θ} + cos θ$ $=sintheta*sintheta/costheta+costheta$

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

$= \frac{1}{cos θ}$ $=1/costheta$

$= sec θ$ $=sectheta$

$= R H S$ $=RHS$

$Q E D$ $QED$

Answer 2 · 2018-03-09T14:02:29

Apply the identities $tan (θ) = \frac{sin θ}{cos θ}$ $tan(theta)=(sin theta)/(cos theta)$ and $sec θ = \frac{1}{cos θ}$ $sec theta=1/(cos theta)$ along with the Pythagorean theorem.

Explanation:

Apply the identity $tan (θ) = \frac{sin θ}{cos θ}$ $tan(theta)=(sin theta)/(cos theta)$ :
$L . H . S . = sin θ \cdot \frac{sin θ}{cos θ} + cos θ$ $L.H.S.=sin theta*sin theta/(cos theta)+cos theta$
$= \frac{{sin}^{2} θ}{cos θ} + cos θ$ $=sin^2 theta/(cos theta)+cos theta$
$= \frac{{sin}^{2} θ + {cos}^{2} θ}{cos θ}$ $=(sin^2 theta+cos^2 theta)/(cos theta)$

Apply the Pythagorean theorem ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta+cos^2 theta=1$
$= \frac{{sin}^{2} θ + {cos}^{2} θ}{cos θ}$ $=(sin^2 theta+cos^2 theta)/(cos theta)$
$= \frac{1}{cos θ}$ $=1/(cos theta)$

By the definition of secants $\frac{1}{cos θ} = sec θ$ $1/(cos theta)=sec theta$ :
$= sec θ$ $=sec theta$
$= R . H . S$ $=R.H.S$

Answer 3 · 2018-03-09T14:06:04

$see explanation$ $"see explanation"$

Explanation:

$using the trigonometric identities$ $"using the "color(blue)"trigonometric identities"$

$∙ x tan θ = \frac{sin θ}{cos θ} and sec θ = \frac{1}{cos θ}$ $•color(white)(x)tantheta=sintheta/costheta" and "sectheta=1/costheta"$

$∙ x {sin}^{2} θ + {cos}^{2} θ = 1$ $•color(white)(x)sin^2theta+cos^2theta=1$

$Consider the left side$ $"Consider the left side"$

$\Rightarrow sin θ tan θ + cos θ$ $rArrsinthetatantheta+costheta$

$= sin θ \times \frac{sin θ}{cos θ} + cos θ$ $=sinthetaxxsintheta/costheta+costheta$

$= \frac{{sin}^{2} θ}{cos θ} + \frac{{cos}^{2} θ}{cos θ} \leftarrow common denominator cos θ$ $=sin^2theta/costheta+cos^2theta/costhetalarr"common denominator "costheta$

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

$= \frac{1}{cos θ} = sec θ = right side \Rightarrow verified$ $=1/costheta=sectheta=" right side "rArr" verified"$

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How do you verify the identify $sin θ tan θ + cos θ = sec θ$ $sinthetatantheta+costheta=sectheta$ ?

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