Question #98419
Show that;sin(90°-theta)/Cosec(90°-theta)+cos(90°-theta)/sec(90°-theta)=1
1
Expert's answer
2019-11-12T11:18:25-0500

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

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

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

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

sin(90θ)cosec(90θ)+cos(90θ)sec(90θ)=cos(θ)sec(θ)+sin(θ)cosec(θ)=cos(θ)1cos(θ)+sin(θ)1sin(θ)=\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)}}=

=cos2(θ)+sin2(θ)=1=cos^2(\theta)+sin^2(\theta)=1


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