CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If ex=1+t1t1+t+1t and tany2=1t1+t then dydx at t=12 is

A
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 12
Let t=cos2θ
ex=1+cos2θ1cos2θ1+cos2θ+1cos2θex=cosθsinθcosθ+sinθex=1tanθ1+tanθ=tan(π4θ)exdxdθ=sec2(π4θ)

tany2=1cos2θ1+cos2θtany2=tanθ12sec2y2dydθ=sec2θ

When t=12=cos2θθ=π6
ex=tan(π4π6)x=lntanπ12
tany2=tanπ6y=π3

So,
dydx=dydθdxdθdydx=2cos2y22sec2θexsec2(π4θ)
Putting t=12,x=lntanπ12,y=π3
dydx=2cos2π62sec2π6elntanπ/12sec2(π12)dydx=2cotπ12sec2(π12)dydx=2tanπ12cos2π12dydx=sinπ6=12

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