Question

Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ

Answer 1

Given:
x = 3cosecθ + 4cotθ              .....(1)
y = 4cosecθ – 3cotθ              .....(2)

Multiplying (1) by 4 and (2) by 3, we get
4x = 12cosecθ + 16cotθ         .....(3) 
3y = 12cosecθ – 9cotθ           .....(4) 

Subtracting (4) from (3), we get
4x − 3y = 25cot θ

⇒ cot θ = \[\frac{4x - 3y}{25}\]

⇒ cot²θ = \[\left( \frac{4x - 3y}{25} \right)^2\]             .....(5)

Multiplying (1) by 3 and (2) by 4, we get
3x = 9cosecθ + 12cotθ          .....(6) 
4y = 16cosecθ – 12cotθ        .....(7) 
Adding (6) and (7), we get
3x + 4y = 25cosecθ

⇒ cosecθ = \[\frac{3x + 4y}{25}\]

⇒ cosec²θ = \[\left(\frac{3x + 4y}{25}\right)^2\]          .....(8)

\[{cosec}^2 \theta - \cot^2 \theta = 1\]

\[{cosec}^2 \theta - \cot^2 \theta = \left( \frac{3x + 4y}{25} \right)^2 - \left( \frac{4x - 3y}{25} \right)^2 = 1\]

\[ \Rightarrow \left( \frac{3x + 4y}{25} \right)^2 - \left( \frac{4x - 3y}{25} \right)^2 = 1\]

\[ \Rightarrow \frac{1}{{25}^2}\left[ \left( 3x + 4y \right)^2 - \left( 4x - 3y \right)^2 \right] = 1\]

\[ \Rightarrow \left( 3x + 4y \right)^2 - \left( 4x - 3y \right)^2 = 625\]