Question in the attachment
Answers
Step-by-step explanation:
Solutions:-
1) Given that
x = a Cos^3 θ----------------(1)
On squaring both sides then
x^2 = a^2 (Cos^3)^2 θ
On raising power to (1/3) both sides then
x^2/3 = a^2/3 Cos^2 θ-----(2)
y = a sin^3 θ------------------(3)
On squaring both sides then
y^2 = a^2 (Sin^3)^2 θ
On raising power to (1/3) both sides then
y^2/3 = a^2/3 Sin^2 θ-------(4)
On adding (2)&(4)
x^2/3 +y^2/3= (a^2/3)Cos^2 θ +( a^2/3) Sin^2 θ
x^2/3+y^2/3= a^2/3[Cos^2 θ +Sin^2 θ]
We know that sin^2A+cos^2A = 1
x^2/3+y^2/3 = a^2/3 (1)
x^2/3 + y^2/3 = a^2/3
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2)Given that
x = Cosec θ - Sin θ-----------(1)
=>x = (1/Sin θ) - Sin θ
=>x = (1-Sin^2θ)/Sinθ
=>x = Cos^2θ/Sinθ
=>x = (Cosθ/Sinθ) Cosθ
=>x = Cot θ Cos θ
On squaring both sides then
=>x^2 = Cot^2 θ Cos^2 θ-----(2)
y=Sec θ - Cos θ
=>y = (1/Cosθ)-Cosθ
=>y=(1-Cos^2 θ)/Cosθ
=>y=Sin^2 θ/Cosθ
=>y = (Sinθ/Cosθ) Sinθ
=>y = Tan θ Sin θ
On squaring both sides then
=>y^2 = Tan^2 θ Sin^2 θ --------(3)
x^2y= (Cot^2 θ Cos^2 θ)×(Tan θ Sin θ)
=>x^2y = Cot^2 θ×tanθ×Cos^2θ×Sinθ
=>x^2y = Cot θ× Cos^2θ×Sinθ
=>x^2y = (Cos θ/Sinθ)×Cos^2θ×Sinθ
=>x^2y = Cos^2 θ× Cos θ
=>x^2y = Cos^3 θ
=>(x^2y)^2/3= (Cos^3 θ)^2/3
=>(x^2y)^2/3= Cos^2 θ ---------------(4)
xy^2 = (Cot θ Cos θ)×(Tan^2 θ Sin^2 θ)
=>xy^2 = Cot Tan^2 θ Sin^2 θ× Cos θ
=>xy^2 = Tan θSin^2 θ× Cos θ
=>xy^2 = (Sin θ/Cos θ)×Sin^2 θ× Cos θ
=>xy^2= Sinθ× Sin^2 θ
=>xy^2 = Sin^3 θ
=>(xy^2)^2/3 = (Sin^3 θ)^2/3
=>(xy^2)^2/3 = Sin^2 θ--------------(5)
On adding (4)&(5)
(xy^2)^2/3 + (x^2y)^2/3 = Sin^2 θ+Cos^2θ
(xy^2)^2/3 + (x^2y)^2/3 = 1
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3) Given that
x = Cot θ + Tan θ
=>x = (Cosθ/Sinθ)+(Sinθ/Cosθ)
=>x= (Cos^2θ+Sin^2θ)/SinθCosθ
=>x= 1/SinθCosθ
=>x = Cosec θ Sec θ
On squaring both sides then
x^2 = Cosec^2 θ Sec^2 θ -----------(1)
y = Sec θ - Cos θ
=>y = (1/Cos θ)-Cosθ
=>y = (1-Cos^2 θ)/Cos θ
=>y = Sin^2 θ/Cos θ
=>y = Tan θ Sin θ
On squaring both sides then
=>y^2 = Tan^2θ Sin^2θ ---------------(2)
x^2y = (Cosec^2 θ Sec^2 θ)(Tan θ Sin θ)
=>x^2y =Cosec^2θ×Sec^2θ×Sin^2θ/Cos θ
=>x^2y = Sec^2 θ /Cos θ
=>x^2y = Sec^2 θ ×Sec θ
=>x^2y = Sec^3 θ
=>(x^2y)^2/3 = (Sec^3 θ )^2/3
=>(x^2y)^2/3 = Sec^2 θ -------------(3)
xy^2 = (Cosec θ Secθ )×(Tan^2 θ Sin^2 θ )
=>xy^2 = Tan^3 θ
=>(xy^2)^2/3 = (Tan^3 θ )^2/3
=>(xy^2)^2/3 = Tan^2 θ -------------(4)
On subtracting (4) from (3)
=>(x^2y)^2/3 - (xy^2)^2/3 = Sec^2 θ - Tan^2 θ
(x^2y)^2/3 - (xy^2)^2/3 = 1
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4) Given that
a= x Sec θ+ y Tan θ
On squaring both sides then
=>a^2 = ( x Sec θ+ y Tan θ)^2
=>a^2 = x^2Sec^2 θ+ y^2 Tan^2 θ+2xy Sec θTan θ
and
b= x Tan θ + y Sec θ
On squaring both sides then
=>b^2 = (x Tan θ+ y Sec θ)^2
=>b^2 = x^2 Tan^2 θ+ y^2 Sec^2θ+2xySecθTanθ
Now
a^2 - b^2
=(x^2Sec^2 θ+ y^2 Tan^2 θ+2xy Sec θTan θ) -
(x^2 Tan^2 θ+ y^2 Sec^2θ+2xySecθTanθ)
=>x^2Sec^2 θ+ y^2 Tan^2 θ- x^2 Tan^2 θ- y^2 Sec^2θ
=>x^2(Sec^2 θ-Tan^2θ)+y^2(Tan^2 θ-Sec^2θ)
=>x^2 (Sec^2 θ-Tan^2θ)-y^2(Sec^2θ-Tan^2 θ)
=>x^2(1)-y^2(1)
=>x^2-y^2
a^2-b^2 = x^2-y^2
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Answers:-
Option (b)
I-S
II-P
III-Q
IV-R
Used formulae:-
- Sin^2 A + Cos^2 A = 1
- Sec^2 A- Tan^2 A = 1
- Cosec^2 A- Cot^2 A= 1
- Tan A = Sin A/Cos A
- Cot A = Cos A/ Sin A
- SecA = 1/CosA
- Cosec A= 1/Sin A
- Tan A = 1/CotA
- Cot A = 1/Tan A
- Tan A× Cot A = 1
- SinA × CosecA = 1
- SecA×CosA =1
Step-by-step explanation:
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