Question

senΘ³÷cosΘ+senΘ.cosΘ ctgΘ

Answer 1

Hola 1) tgθ / (1-ctg θ) = (sinθ/cosθ) / (1-(cosθ/sinθ)) tgθ / (1-ctg θ) = (sinθ/cosθ) / (sinθ-cosθ)/sinθ) tgθ / (1-ctg θ) = sin^2θ/ ((cosθ)(sinθ-cosθ)) ctgθ / (1-tg θ) = (cosθ/sinθ) / (1-(sinθ/cosθ)) ctgθ / (1-tg θ) = (cosθ/sinθ) / ((-1)(sinθ-cosθ)/cosθ) ctgθ / (1-tg θ) = -cos^2θ / ((sinθ) (sinθ-cosθ) [ tgθ / (1-ctg θ) ] + [ ctg θ / (1-tg θ) ] = = sin^2θ/ ((cosθ)(sinθ-cosθ)) - cos^2θ / ((sinθ) (sinθ-cosθ) = (1/(sinθ-cosθ)) ( (sin^2θ/cosθ) - (cos^2θ/sinθ) ) = (1/(sinθ-cosθ)) ( (sin^3θ - cos^3θ)/(sinθ cosθ)) Aplicamos factoreo de diferencia de cubos = ( (sin^2θ + sinθ cosθ + cos^2θ)/(sinθ cosθ)) = (sinθ/cosθ) + 1 + (cosθ/sinθ) = 1 + tgθ + ctgθ 

Answer 2

ahi te va la solución.........................