Answer to Question #98419 in Trigonometry for Shams

Question

Show that;sin(90°-theta)/Cosec(90°-theta)+cos(90°-theta)/sec(90°-theta)=1

Answer 1

$sin(90^\circ-\theta)=cos(\theta)$

$cos(90^\circ-\theta)=sin(\theta)$

$sec (90^\circ-\theta)=\frac{1}{cos(90^\circ-\theta)}=\frac{1}{sin(\theta)}$

$Cosec (90^\circ-\theta)=\frac{1}{sin(90^\circ-\theta)}=\frac{1}{cos(\theta)}$

$\frac{sin(90^\circ-\theta)}{cosec(90^\circ-\theta)}+\frac{cos(90^\circ-\theta)}{sec(90^\circ-\theta)}=\frac{cos(\theta)}{sec(\theta)}+\frac{sin(\theta)}{cosec(\theta)}=\frac{cos(\theta)}{\frac{1}{cos(\theta)}}+\frac{sin(\theta)}{\frac{1}{sin(\theta)}}=$

$=cos^2(\theta)+sin^2(\theta)=1$