Question

If x = sec 0 + tan 0 and y = sec 0
sec 0 – tan 8, then​

Answer 1

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: