Eliminate θ If 2x= 3- 4cotθ, 3y = 5+ 3cosecθ
Answers
Given :-
2x = 3 - 4 cot θ
3y = 5 + 3 cosec θ
To find :-
Eliminate θ
Solution :-
Given equations are :
2x = 3 - 4 cot θ ---------(1)
=> 2x-3 = -4 cot θ
=> (2x-3)/-4 = cot θ
=> cot θ = -(2x-3)/4
On squaring both sides then
=> (cot θ)² = [-(2x-3)/4]²
=> cot² θ = [(2x)²-2(2x)(3)+3²]/16
Since, (a-b)² = a²-2ab+b²
Where, a = 2x and b = 3
=> cot² θ = (4x²-12x+9)/16 ---------------(2)
3y = 5 + 3 cosec θ --------(3)
=> 3y-5 = 3 cosec θ
=> (3y-5)/3 = cosec θ
=> cosec θ = (3y-5)/3
On squaring both sides then
=> (cosec θ)² = [(3y-5)/3]²
=> cosec² θ = [(3y)²-2(3y)(5)+5²]/9
Since, (a-b)² = a²-2ab+b²
Where, a = 3y and b = 5
=> cosec² θ = (9y²-30y+25)/9 -----------(4)
On subtracting (2) from (4) then
cosec² θ-cot² θ
= [(9y²-30y+25)/9]-[(4x²-12x+9)/16]
We know that
cosec² A - cot² A = 1
Therefore,
1 = [(9y²-30y+25)/9]-[(4x²-12x+9)/16]
=> [(9y²-30y+25)/9]-[(4x²-12x+9)/16] = 1
LCM of 9 and 16 = 144
[16(9y²-30y+25)-9(4x²-12x+9)] /144 = 1
=> [16(9y²-30y+25)-9(4x²-12x+9)] = 1×144
=> [16(9y²-30y+25)-9(4x²-12x+9)] = 144
=> (144y²-480y+400)-(36x²-108x+81) = 144
=> 144y²-480y+400-36x²+108x-81 = 144
=> 144y²-36x²-480y+108x+319 = 144
=> 144y²-36x²-480y+108x +319-144 = 0
=> 144y²-36x²-480y+108x+175 = 0
or
=> -36x²+108x-480y+144y²+175 = 0
=> 36x²-108x+480y-144y²-175 = 0
Answer :-
The answer for the given problem after eliminating θ is 144y²-36x²-480y+108x+175 =0 (or)
36x²-108x+480y-144y²-175 = 0
Used formulae:-
→ (a-b)² = a²-2ab+b²
→ cosec² A - cot² A = 1
Step-by-step explanation:
Given :-
2x = 3 - 4 cot θ
3y = 5 + 3 cosec θ
To find :-
Eliminate θ
Solution :-
Given equations are :
2x = 3 - 4 cot θ ---------(1)
=> 2x-3 = -4 cot θ
=> (2x-3)/-4 = cot θ
=> cot θ = -(2x-3)/4
On squaring both sides then
=> (cot θ)² = [-(2x-3)/4]²
=> cot² θ = [(2x)²-2(2x)(3)+3²]/16
Since, (a-b)² = a²-2ab+b²
Where, a = 2x and b = 3
=> cot² θ = (4x²-12x+9)/16 ---------------(2)
3y = 5 + 3 cosec θ --------(3)
=> 3y-5 = 3 cosec θ
=> (3y-5)/3 = cosec θ
=> cosec θ = (3y-5)/3
On squaring both sides then
=> (cosec θ)² = [(3y-5)/3]²
=> cosec² θ = [(3y)²-2(3y)(5)+5²]/9
Since, (a-b)² = a²-2ab+b²
Where, a = 3y and b = 5
=> cosec² θ = (9y²-30y+25)/9 -----------(4)
On subtracting (2) from (4) then
cosec² θ-cot² θ
= [(9y²-30y+25)/9]-[(4x²-12x+9)/16]
We know that
cosec² A - cot² A = 1
Therefore,
1 = [(9y²-30y+25)/9]-[(4x²-12x+9)/16]
=> [(9y²-30y+25)/9]-[(4x²-12x+9)/16] = 1
LCM of 9 and 16 = 144
[16(9y²-30y+25)-9(4x²-12x+9)] /144 = 1
=> [16(9y²-30y+25)-9(4x²-12x+9)] = 1×144
=> [16(9y²-30y+25)-9(4x²-12x+9)] = 144
=> (144y²-480y+400)-(36x²-108x+81) = 144
=> 144y²-480y+400-36x²+108x-81 = 144
=> 144y²-36x²-480y+108x+319 = 144
=> 144y²-36x²-480y+108x +319-144 = 0
=> 144y²-36x²-480y+108x+175 = 0
or
=> -36x²+108x-480y+144y²+175 = 0
=> 36x²-108x+480y-144y²-175 = 0