Question

3. Verifica las siguientes identidades: a) tan θcotθ − cos2θ = sen2θ b) secθ cscθ + sen θ cosθ = 2 tan θ c) (csc α + cot α)(csc α − cot α) = 1 d) 1 − sen2β 1+cosβ = cos β e) tan θ−cotθ tan θ+cotθ = sen2θ − cos2θ f) cos2α−sen2α 1−tan2α = cos 2α

Answer 1

Al verificar las identidades trigonométricas se obtiene:

$a)tan\theta cot\theta - cos^{2}\theta = sen^{2}\theta$

Aplicar identidades trigonométricas;

$tan\theta = \frac{sen\theta}{cos\theta}\\cos^{2}\theta + sen^{2}\theta = 1\\cot\theta = \frac{cos\theta}{sen\theta}$

$=tan\theta cot\theta - cos^{2}\theta \\= \frac{sen\theta}{cos\theta}.\frac{cos\theta}{sen\theta} - cos^{2}\theta \\= 1 - cos^{2}\theta\\=sen^{2}\theta$

$b)\frac{sec\theta}{csc\theta}+\frac{sen\theta}{cos\theta} = 2tan\theta$

Aplicar identidades trigonométricas;

$sec\theta = \frac{1}{cos\theta}\\csc\theta= \frac{1}{sen\theta}\\tan\theta = \frac{sentheta}{cos\theta}$

$=\frac{sec\theta}{csc\theta}+\frac{sen\theta}{cos\theta} \\=\frac{\frac{1}{cos\theta} }{\frac{1}{sen\theta} }+ \frac{sen\theta}{cos\theta} \\=\frac{sen\theta}{cos\theta} +\frac{sen\theta}{cos\theta} \\= tan\theta + tan\theta\\= 2tan\theta$

$c)(csc\alpha + cot\alpha) (csc\alpha - cot\alpha) = 1$

Aplicar una diferencia de cuadrados;

(a + b)(a - b) = a² - b²

Aplicar identidades trigonométricas;

$csc^{2}\alpha = 1+ cot^{2}\alpha$

$=(csc\alpha + cot\alpha) (csc\alpha - cot\alpha) \\= csc^{2} \alpha - cot^{2} \alpha\\=(1+cot^{2} \alpha) - cot^{2} \alpha\\= 1 + cot^{2} \alpha - cot^{2} \alpha\\= 1$

$d)1-\frac{sen^{2}\beta}{1+cos\beta }=cos\beta$

Aplicar identidad trigonométrica;

$cos^{2}\theta + sen^{2}\theta = 1$

$=1-\frac{sen^{2}\beta}{1+cos\beta }\\=\frac{1 + cos\beta - (1-cos^{2}\beta)}{1 + cos\beta} \\=\frac{1 + cos\beta - 1+cos^{2}\beta}{1 + cos\beta}\\=\frac{cos\beta +cos^{2}\beta}{1 + cos\beta}\\= \frac{cos\beta(1 +cos\beta)}{1 + cos\beta}\\=cos\beta$

$e) \frac{tan\theta - cot\theta}{tan\theta +cot\theta}=sen^{2}\theta - cos^{2}\theta$

Aplicar identidad trigonométrica;

$tan\theta = \frac{sen\theta}{cos\theta}\\cot\theta = \frac{cos\theta}{sen\theta}\\cos^{2}\theta + sen^{2}\theta = 1$

$= \frac{tan\theta - cot\theta}{tan\theta +cot\theta}\\=\frac{\frac{sen\theta}{cos\theta} - \frac{cos\theta}{sen\theta} }{\frac{sen\theta}{cos\theta} + \frac{cos\theta}{sen\theta}} \\=\frac{\frac{sen^{2} \theta - cos^{2}\theta }{cos\theta sen\theta}}{\frac{sen^{2} \theta cos^{2}\theta }{cos\theta sen\theta} } \\=\frac{sen^{2} \theta - cos^{2}\theta}{1} \\= sen^{2} \theta - cos^{2}\theta$

$f) \frac{cos^{2}\alpha -sen^{2}\alpha}{1-tan^{2}\alpha} = cos^{2}\alpha$

Aplicar identidad trigonométrica;

$tan^{2} \theta = \frac{sen^{2}\theta}{cos^{2}\theta}\\cos^{2}\theta + sen^{2}\theta = 1$

$=\frac{cos^{2}\alpha -sen^{2}\alpha}{1-tan^{2}\alpha} \\=\frac{cos^{2}\alpha -sen^{2}\alpha}{1-\frac{sen^{2}\alpha}{cos^{2}\alpha} }\\=\frac{cos^{2}\alpha -sen^{2}\alpha}{\frac{cos^{2}\alpha-sen^{2}\alpha}{cos^{2}\alpha} }\\=\frac{cos^{2}\alpha (cos^{2}\alpha -sen^{2}\alpha)}{cos^{2}\alpha-sen^{2}\alpha}\\=cos^{2}\alpha$