Check the solution to "Let x=asin⁡θ in a2−x2. Then find cos⁡θ tan⁡θ"

High School
Geometry
Trigonometry
Trigonometric Functions

Lipossig

Answered question

2021-01-24

Let $x = a \sin θ i n \sqrt{a^{2} - x^{2}}$ . Then find $\cos θ \tan θ$

Answer & Explanation

Gennenzip

Skilled2021-01-25Added 96 answers

$\sqrt{a^{2} - s^{2} \sin^{2} (θ)} =$
Factor out common term $a^{2}$
$= \sqrt{a^{2} (1 - \sin^{2} (θ))}$
Apply radical rule $\sqrt[n]{a b} = \sqrt[n]{a} \sqrt[n]{b}$ , assuming $\frac{a \geq 0}{b \geq 0}$
$= \sqrt{a^{2}} \sqrt{- \sin^{2} (θ) + 1}$
Apply radical rule $\sqrt[n]{a^{n}} = a$ , assuming $a \geq 0$ Use the following identity: $\cos^{2} (x) + \sin^{2} (x) = 1$
Therefore $1 - \sin^{2} (x) = \cos^{2} (x)$
$= a \sqrt{\cos^{2} (θ)} = a \cos (θ)$
Therefore $\sqrt{a^{2} - a^{2} \sin^{2} (θ)} = a \cos (θ)$
$⟹ \cos θ = \sqrt{a^{2} - a^{2}} \frac{\sin^{2} (θ)}{a}$
$⟹ \frac{a \sin (θ)}{\sqrt{a^{2} - a^{2} \sin^{2} (θ)}} = \tan (θ)$
$\sqrt{a^{2} - a^{2} \sin^{2} (θ)} = a \cos (θ)$

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