# If 3 cosθ = 5 sinθ, then the value of (5 sinθ - 2sec^3θ + 2 cosθ)/(5sinθ + 2sec^3θ - 2cosθ)

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### 2 Answers

You should divide by cos theta such that:

`5sin theta/cos theta = 3`

`cos theta = (5/3)sin theta`

`cos^2 theta = (25/9)sin^2 theta`

Using `cos^2 theta = 1 - sin^2 theta` yields:

`1 - sin^2 theta = 25/9 sin^2 theta => (34/9)sin^2 theta = 1`

`sin^2 theta = 9/34 => cos^2 theta = 1 - 9/34 => cos^2 theta = 25/34 => cos^4 theta = 625/1156`

You need to remember that `sec theta = 1/cos theta` , hence, `sec^3 theta = 1/(cos^3 theta)`

Factoring out `cos theta` to numerator and denominator yields:

`(cos theta(3 - 2/(cos^4 theta) + 2))/(cos theta(3+ 2/(cos^4 theta)- 2))= (5 - 2/(cos^4 theta))/(1 + 2/(cos^4 theta))`

`((5cos^4 theta - 2)/(cos^4 theta))/((cos^4 theta + 2)/(cos^4 theta)) = (5cos^4 theta - 2)/(cos^4 theta + 2)`

Substituting `625/1156` for `cos^4 theta` yields:

`(5cos^4 theta - 2)/(cos^4 theta + 2) = (5*625/1156 - 2)/(625/1156 + 2) = (3125-2312)/(625 + 2312)`

`(5cos^4 theta - 2)/(cos^4 theta + 2) = 813/2937`

**Hence, evaluating the given expression under the given condition `3 cos theta = 5 sin theta` yields `(5sin theta - 2sec^3 theta + 2 cos theta)/(5sin theta+ 2sec^3 theta- 2 cos theta) = (5cos^4 theta - 2)/(cos^4 theta + 2) = 813/2937.` **

`3 costheta = 5 sintheta`

`sintheta/costheta = 3/5`

`tantheta = 3/5`

`sec^2theta = 1+tan^2theta = 1+(3/5)^2 = 34/25`

`sectheta = sqrt34/5`

`costheta = 1/sectheta = 5/sqrt34`

`cosec^2theta = 1+cot^2theta = 1+1/tan^2theta = 1+(5/3)^2 = 34/9`

`cosectheta = sqrt34/3`

`sintheta =1/(cosectheta) = 3/sqrt34`

`(5 sintheta - 2sec^3theta + 2costheta)/(5sintheta + 2sec^3theta - 2costheta)`

`= (5*3/sqrt34-2(sqrt34/5)^3+2*5/sqrt34)/(5*3/sqrt34+2*(sqrt34/5)^3-2*5/sqrt34)`

`= [(125*15-2*34*34+10*125)/(125*sqrt34)]/[(15*125+2*34*34-10*125)/(125*sqrt34)]`

`= 271/979`

Therefore;

`(5 sintheta - 2sec^3theta + 2costheta)/(5sintheta + 2sec^3theta - 2costheta) = 271/979`

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