If x = sec 0 + tan 0 and y = sec 0
sec 0 – tan 8, then
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Answer:
x=secθ−cosθ
x
2
−4=(secθ−cosθ)
2
+4
=(secθ+cosθ)
2
y=sec
19
θ−cos
n
θ
y
2
+4=(sec
n
θ−cos
n
θ)
2
+4
⇒ (sec
2
θ+cos
n
θ)
2
dθ
dy
=nsec
n−1
θ(secθ+tanθ)−ncos
n−1
θ(−sinθ)
⇒ ntanθ(sec
n
θ+cos
n
θ)
dθ
dy
=secθtanθsinθ
=tanθ(secθ+cosx)
∴
dfracdydx=
tanθ(secθ+cosθ)
ntanθ(sec
n
θ+cos
n
θ)
(
dx
dy
)
2
=n
2
(x
2
+4)
(y
2
+4)
∴ (x
2
+4)(
dx
dy
)
2
=n
2
(4+y
2
Step-by-step explanation:
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