wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Match the following columns with the values obtained for the solution.

I.
xcosθ+ysinθ=a,
xsinθycosθ=b

a) (x2y2)2=16xy
II.
x=secθ+tanθ,
y=secθtanθ
b) xy=1
III.
xsecθ+ytanθ=a,
xtanθ+ysecθ=b

c) x2y2=a2b2

IV.
x=cotθ+cosθ,
y=cotθcosθ

d) x2+y2=a2+b2

A
Ic,IId,IIIa,IVb
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Id,IIb,IIIc,IVa
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Ia,IIc,IIId,IVb
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Ib,IIc,IIIa,IVd
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B Id,IIb,IIIc,IVa
I. xcosθ+ysinθ=a.....(i)
xsinθycosθ=b.....(ii)
Squaring and adding equations (i) and (ii), we get
[xcosθ+ysinθ]2+[xsinθycosθ]2=a2+b2
x2cos2θ+y2sin2θ+2xysinθcosθ+x2sin2θ+y2cos2θ2xysinθcosθ=a2+b2
x2[sin2θ+cos2θ]+y2[sin2θ+cos2θ]=a2+b2
x2+y2=a2+b2
which is (d).
II. x=secθ+tanθ
y=secθtanθ
xy=(secθ+tanθ)(secθtanθ)
xy=sec2θtan2θ
xy=1[1+tan2θ=sec2θ]
which is (b).
III. xsecθ+ytanθ=a ....... (iii)
xtanθ+ysecθ=b ......... (iv)
Squaring and then subtracting equation (iii) from (iv), we get
[x2sec2θ+y2tan2θ+2xysecθtanθ][x2tan2θ+y2sec2θ+2xysecθtanθ]=a2b2
x2[sec2θtan2θ]y2[sec2θtan2θ]=a2b2
x2y2=a2b2
which is (c).
IV. x=cotθ+cosθ
y=cotθcosθ
Squaring both equations, we get
x2=cot2θ+cos2θ+2cotθcosθ ......... (v)
y2=cot2θ+cos2θ2cotθcosθ ......... (vi)
Subtract equation (vi) from (v), we get
x2y2=4cotθcosθ
Squaring both sides, we get
(x2y2)2=16cot2θcos2θ ......... (vii)
Now, xy=(cotθ+cosθ)(cotθcosθ)
xy=cot2θcos2θ
xy=cos2θsin2θ[1sin2θ]
xy=cot2θcos2θ ......... (viii)
Substitute (viii) in equation (vii), we get
(x2y2)2=16xy
which is (a).
Therefore, I-d, II-b, III-c, IV-a

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Trigonometric Identities_Concept
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon