Question

Ayuda con estas identidades trigonométricas es para mañana .
* 1/tan^3x + 1 = csc^2x
* cot^2x + 1/tanx cotx = csc^2x
* cot^2x + 1/cosx secx = csc^2x
*tan^2x + 1/senx cscx = sec^2x
*cot^2x + sen^2x + cos^2x = csc^2x

Answer 1

Primero:
$\frac{1}{tan^2 \alpha } +{1}= \\ {ctg^2 \alpha}+{1}= \\ { \frac{cos^2 \alpha }{sen^2 \alpha }} +{1}= \\ {\frac{cos^2 \alpha +sen^2 \alpha }{sen^2}=} \\ { \frac{1}{sen^2 \alpha }= } \\ {csc^2 \alpha }}$

S egunda:
${ctg^2 \alpha }+ \frac{1}{tan \alpha ctg \alpha }=} \\ { \frac{cos^2 \alpha }{sen^2 \alpha} +{1}=} \\ \frac{cos^2 \alpha+ sen^2 \alpha }{sen^2 \alpha} =} \\ \frac{1}{sen^2 \alpha} =} \\ {csc^2 \alpha }$

La tercera:
${ctg^2 \alpha }+ \frac{1}{cos \alpha sec \alpha }=} \\ { \frac{cos^2 \alpha }{sen^2 \alpha} +{1}=} \\ \frac{cos^2 \alpha+ sen^2 \alpha }{sen^2 \alpha} =} \\ \frac{1}{sen^2 \alpha} =} \\ {csc^2 \alpha }$

La cuarta:
${tan^2 \alpha }+ \frac{1}{sen \alpha csc \alpha }=} \\ { \frac{sen^2 \alpha }{cos^2 \alpha} +{1}=} \\ \frac{sen^2 \alpha+ cos^2 \alpha }{cos^2 \alpha} =} \\ \frac{1}{cos^2 \alpha} =} \\ {sec^2 \alpha }$

La última:
${ctg^2 \alpha }+sen^2 \alpha+ cos^2 \alpha }=} \\ { \frac{cos^2 \alpha }{sen^2 \alpha} +{1}=} \\ \frac{cos^2 \alpha+ sen^2 \alpha }{sen^2 \alpha} =} \\ \frac{1}{sen^2 \alpha} =} \\ {csc^2 \alpha }$

Salu2.

Suerte... :)