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MATU • Trigonometria
MATU_TREC_042
Compendio de Trigonometría
Enunciado
Que valor de $A$ que satisface la identidad:
$$\frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = \frac{1}{\cos^2 x - A}$$
$$\frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x} = \frac{1}{\cos^2 x - A}$$
Solución Paso a Paso
1. Simplificación del LHS:
$$\frac{(1 + \tan x)^2 + (1 - \tan x)^2}{(1 - \tan x)(1 + \tan x)} = \frac{1 + 2 \tan x + \tan^2 x + 1 - 2 \tan x + \tan^2 x}{1 - \tan^2 x}$$
$$LHS = \frac{2(1 + \tan^2 x)}{1 - \tan^2 x}$$
Usando $\tan^2 x = \frac{sen^2 x}{\cos^2 x}$:
$$LHS = \frac{2(1 + \frac{sen^2 x}{\cos^2 x})}{1 - \frac{sen^2 x}{\cos^2 x}} = \frac{2(\frac{\cos^2 x + sen^2 x}{\cos^2 x})}{\frac{\cos^2 x - sen^2 x}{\cos^2 x}} = \frac{2(1)}{\cos^2 x - sen^2 x}$$
2. Transformación para hallar A:
Queremos que $\frac{2}{\cos^2 x - sen^2 x} = \frac{1}{\cos^2 x - A}$.
Invirtiendo: $\frac{\cos^2 x - sen^2 x}{2} = \cos^2 x - A$
$\cos^2 x - sen^2 x = 2 \cos^2 x - 2A$
$2A = \cos^2 x + sen^2 x = 1$
$A = 1/2$.
$$\frac{(1 + \tan x)^2 + (1 - \tan x)^2}{(1 - \tan x)(1 + \tan x)} = \frac{1 + 2 \tan x + \tan^2 x + 1 - 2 \tan x + \tan^2 x}{1 - \tan^2 x}$$
$$LHS = \frac{2(1 + \tan^2 x)}{1 - \tan^2 x}$$
Usando $\tan^2 x = \frac{sen^2 x}{\cos^2 x}$:
$$LHS = \frac{2(1 + \frac{sen^2 x}{\cos^2 x})}{1 - \frac{sen^2 x}{\cos^2 x}} = \frac{2(\frac{\cos^2 x + sen^2 x}{\cos^2 x})}{\frac{\cos^2 x - sen^2 x}{\cos^2 x}} = \frac{2(1)}{\cos^2 x - sen^2 x}$$
2. Transformación para hallar A:
Queremos que $\frac{2}{\cos^2 x - sen^2 x} = \frac{1}{\cos^2 x - A}$.
Invirtiendo: $\frac{\cos^2 x - sen^2 x}{2} = \cos^2 x - A$
$\cos^2 x - sen^2 x = 2 \cos^2 x - 2A$
$2A = \cos^2 x + sen^2 x = 1$
$A = 1/2$.