If $tan A + sec A = 2$ $tanA+secA=2$ then find $cos A$ $cosA$ ?

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If $tan A + sec A = 2$ $tanA+secA=2$ then find $cos A$ $cosA$ ?

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Answer 1 · 2016-10-27T21:58:35

$cos A = \frac{4}{5}$ $cosA=4/5$

Explanation:

$tan A + sec A = 2$ $tanA+secA=2$

Substituting the definition of $tan$ $tan$ and $sec$ $sec$ gives us:
$\frac{sin A}{cos A} + \frac{1}{cos A} = 2$ $sinA/cosA+1/cosA=2$

Multiplying by $cos A$ $cosA$ :
$sin A + 1 = 2 cos A$ $sinA+1=2cosA$

$:. sinA=2cosA - 1$
$:. sin^2A=(2cosA - 1)^2$

Using the fundamental identity $sin^2X+cos^2X-=1$
$1-cos^2A=(2cosA - 1)^2$
$:. 1-cos^2A=4cos^2A-4cosA+1$
$:. 5cos^2A-4cosA = 0$
$:. cosA(5cosA-4) = 0$
$cosA=0$ or $5cosA-4 = 0 => cosA=4/5$

We were told that A is an acute angle, so we can eliminate the solution $cosA=0$

Hence, $cosA=4/5$

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If $tan A + sec A = 2$ $tanA+secA=2$ then find $cos A$ $cosA$ ?

1 Answer

Explanation:

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