Cot2A+tanA=?

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Cot2A+tanA=?

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Answer 1 · 2018-02-24T18:24:58

$csc 2 A$ $csc2A$

Explanation:

$cot 2 A + tan A$ $cot2A+tanA$

$= \frac{cos 2 A}{sin 2 A} + \frac{sin A}{cos A}$ $=(cos2A)/(sin2A)+sinA/cosA$

$= \frac{cos 2 A cos A + sin 2 A sin A}{sin 2 A cos A}$ $=(cos2AcosA+sin2AsinA)/(sin2AcosA)$

$= \frac{cos (2 A - A)}{sin 2 A cos A}$ $=cos(2A-A) /(sin2AcosA)$

$= \frac{cos A}{sin 2 A cos A}$ $=cosA/(sin2AcosA)$

$= \frac{1}{sin 2 A}$ $=1/(sin2A)$

$= csc 2 A$ $=csc2A$

Formulae:

$cos C cos D + sin C sin D = cos (C - D)$ $cosCcosD+sinCsinD=cos(C-D)$
$cos 2 A cos A + sin 2 A sin A = cos (2 A - A)$ $cos2AcosA+sin2AsinA=cos(2A-A)$

Answer 2 · 2018-02-24T18:34:57

= csc 2A

Explanation:

Call tan A = t , and apply the trig identity
$tan 2 A = \frac{2 tan A}{1 - {tan}^{2} A}$ $tan 2A = (2tan A)/(1 - tan^2 A)$
The expression becomes:
$f (A) = cot 2 A + tan A = \frac{1 - t^{2}}{2 t} + t = \frac{(1 - t^{2}) + 2 t^{2}}{2 t}$ $f (A) = cot 2A + tan A = (1 - t^2)/(2t) + t = ((1 - t^2) + 2t^2)/(2t)$
$f (A) = \frac{1 + t^{2}}{2 t} = \frac{1 + {tan}^{2} A}{2 tan A}$ $f(A) = (1 + t^2)/(2t) = (1 +tan^2 A)/(2tan A)$
Using trig identity, replace $(1 + {tan}^{2} A)$ $(1 + tan^2 A)$ by $({sec}^{2} A)$ $(sec^2 A)$ , we get:
$f (A) = ({sec}^{2} A) (\frac{2 sin A}{cos A}) = \frac{cos A}{2 sin A . {cos}^{2} A}$ $f(A) = (sec^2 A)((2sin A)/(cos A)) = (cos A)/(2sin A.cos^2 A)$
Note that: $sin 2 A = 2 sin A . cos A$ $sin 2A = 2sin A.cos A$ -->
$f (A) = \frac{1}{sin 2 A} = csc 2 A$ $f(A) = 1/(sin 2A) = csc 2A$

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Cot2A+tanA=?

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