Question

If x= secθ-cosθ,y=secⁿθ-cosⁿθ,then......,Select Proper option from the given options.
(a) (x²+4)(dy/dx)²=n²(y²+4)
(b) (x²-4)(dy/dx)²=n²(y²-4)
(c) (x²+4)(dy/dx)²=1
(d) (x²+4)(dy/dx)²=y²+4

Answer 1

Hello,

Answer: (a) (x²+4)(dy/dx)²=n²(y²+4)
Solution:


x= secθ-cosθ

dx/dθ = secθ tanθ - (-sin θ)

dx/dθ = secθ tanθ +sin θ

dx/dθ = tanθ (secθ + cos θ) ---eq1

same way differentiate y =secⁿθ-cosⁿθ

dy/dθ = n sec(ⁿ-1)θ secθ tanθ - n cos(ⁿ-1)θ(-sinθ)

dy/dθ = n secⁿθ tanθ + n cosⁿθ (tanθ)

dy/dθ = n tanθ (secⁿθ + cosⁿθ ) -----eq2

to find dy/dx : calculate dy/dθ *dθ/dx

so,
dy/dx = [n tanθ (secⁿθ + cosⁿθ )]/[ tanθ (secθ + cos θ)]

dy/dx = [n (secⁿθ + cosⁿθ )]/[ (secθ + cos θ)]

(dy/dx)² = [n^2 (secⁿθ + cosⁿθ )^2]/[ (secθ + cos θ)^2]

= [n^2 (sec^2ⁿθ + cos^2ⁿθ +2secⁿθcosⁿθ )]/[ (sec^2θ + cos ^2θ + 2secθcos θ)]

(dy/dx)²= [n^2 (sec^2ⁿθ + cos^2ⁿθ +2)]/[ (sec^2θ + cos ^2θ + 2)] -----eq3

now find
${x}^{2}$
x²= (secθ - cosθ)²

x² = sec^2θ + cos^2θ -2
if now add 4 in LHS then we get sec^2θ + cos^2θ +2 in RHS

(x²+4) = sec^2θ + cos^2θ +2 ----eq4

By the same way

(y²+4) = sec^(2n)θ + cos^(2n)θ +2 ----eq5

Now place these values from eq4 and eq 5 in eq3

[ (sec^2θ + cos ^2θ + 2)](dy/dx)²= [n^2 (sec^2ⁿθ + cos^2ⁿθ +2)]

(x²+4)(dy/dx)^2 =n²(y²+4)

Hope it helps you.