Question

x=a(cost+log tan t/2), y= asint,Find dy/dx :(Wherever y is defined as a function of x and dx/dt or dx/dθ≠0)

Answer 1

x = a(cost + log tant/2)

differentiate with respect to t,

dx/dt = d{a(cost + logtan t/2)}/dt

= a[d(cost)/dt + d(logtan t/2)/dt]

= a[ -sint + 1/tan t/2 × sec²t/2× 1/2]

= a [ -sint + 1/{2sin t/2.cos t/2} ]

we know, 2sinA.cosA = sin2A , use it here

= a [ -sint + 1/sint]

= a { 1 - sin²t}/sint

= acos²t/sint

= acott. cost .......(1)

again, y = asint

dy/dt = acost .......(2)

now, dy/dx = {dy/dt}/{dx/dt}

= {acost}{acott.cost}

= 1/cott

hence, dy/dx = tan t

Answer 2

HELLO DEAR,



GIVEN:-
x = a(cost + log tant/2)

differentiate x w.r.t t

dx/dt = d{a(cost + logtan t/2)}/dt

= a[d(cost)/dt + d(logtan t/2)/dt]

= a[ -sint + 1/2 × 1/tan(t/2) × sec²(t/2)]

= a[ -sint + 1/{2sin t/2.cos t/2} ]

we know, sin2x = 2sinxcosx

= a[ -sint + 1/sint]

= a{ 1 - sin²t}/sint

= acos²t/sint

= acot t . cost----------( 1 )


AND,
y = asint

dy/dt = acost ------------( 2 )

now,
divide-----( 1 ) by -----( 2 )

dy/dx = {dy/dt}/{dx/dt}

=> {acost}/{acot t . cost}

=> 1/cot t

=> dy/dx = tan t


I HOPE ITS HELP YOU DEAR,
THANKS