Question

Eliminate theta from x = cos theta + sin theta, y = cos theta . sin theta

Answer 1

Answer:

Squaring both sides of q tan θ + p sec θ = x we get, 

(q tan θ + p sec θ)2 = x2 , …………….. (A) 

Now, squaring both sides of p tan θ + q sec θ = y we get, 

(p tan θ + q sec θ)2 = y2, …………….. (B) 

Now subtract (B) from (A) we get, 

x2 - y2 = (q tan θ + p sec θ)2 - (p tan θ + q sec θ) 2

⇒ x2 - y2 = (q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ) - (p2 tan2 θ + q2 sec2 θ + 2pq tan θ sec θ) 

⇒ x2 - y2 = q2 tan2 θ + p2 sec2 θ + 2qp tan θ sec θ - p2 tan2 θ - q2 sec2 θ - 2pq tan θ sec θ

⇒ x2 - y2 = q2 tan2 θ - p2 tan2 θ + p2 sec2 θ - q2sec2 θ

⇒ x2 - y2 = tan2 θ (q2 – p2) + sec 2 θ (p2 - q2)

⇒ x2 - y2 = - tan2 θ (p2 - q2) + sec 2 θ (p2 - q2) ⇒ x2 - y2 = sec2 θ (p2 - q2) - tan2 θ (p2 - q2)

⇒ x2 - y2 = (p2 – q2) (sec2 θ - tan2 θ) 

⇒ x2 - y2 = (p2 – q2)(1), [Since sec 2 θ - tan2 θ = 1] 

⇒ x2 - y2 = p2 – q2

Hence the required eliminant is x2 - y2 = p2 - q2.

hope it will help u

Answer 2

Answer:

After $\theta$ elimination answer is ${x}^{2} - 2y = 1$

Step-by-step explanation:

Given,

$x = \cos(\theta) + \sin(\theta)$ and $y = \cos(\theta) \sin(\theta)$

Here we want to eliminate $\theta$

Now we know

${sin}^{2} \theta + {cos}^{2} \theta = 1$

${( \sin\theta + \cos\theta) }^{2} - 2sin\theta \: cos\theta = 1$

Here value of $\cos(\theta) \sin(\theta)$ is y and value of $\cos(\theta) + \sin(\theta)$ is x.

So,

${( \sin\theta + \cos\theta) }^{2} - 2sin\theta \: cos\theta = 1 \\ {x}^{2} - 2y = 1$

After $\theta$ elimination, answer is ${x}^{2} - 2y = 1$

Here applied formula is ${a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab$

Some extra important formulas of Trigonometry,

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$

know more about Trigonometry,

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