# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.6

### Question 1. x = 2at2, y = at4

Solution:

Here, x = 2at2, y = at

= 2a

= 2a (2t)

= 4at

And, now

= a

= a (4t3)

= 4at3

Now, as

= t2

### Question 2. x = a cos(Î¸), y = b cos(Î¸)

Solution:

Here, x = a cos(Î¸), y = b cos(Î¸)

= a

= a (-sin(Î¸))

= – a sin(Î¸)

And, now

= b

= b (-sin(Î¸))

= – b sin(Î¸)

Now, as

### Question 3. x = sin(t), y = cos(2t)

Solution:

Here, x = sin(t), y = cos(2t)

= cos(t)

And, now

= -sin(2t)

= – 2sin(2t)

Now, as

(Using the identity: sin(2Î¸) = 2 sinÎ¸ cosÎ¸)

= – 4 sin(t)

### Question 4. x = 4t, y =

Solution:

Here, x = 4t, y = 4/t

= 4

= 4

And, now

= 4

= 4

= 4

= 4

Now, as

### Question 5. x = cos(Î¸) â€“ cos(2Î¸), y = sin(Î¸) â€“ sin(2Î¸)

Solution:

Here, x = cos(Î¸) â€“ cos(2Î¸), y = sin(Î¸) â€“ sin(2Î¸)

= – sin(Î¸) – (-sin(2Î¸))

= – sin(Î¸) + 2sin(2Î¸)

And, now

= cos(Î¸) – (cos(2Î¸))

= cos(Î¸) – (2 cos(2Î¸)

Now, as

### Question 6. x = a (Î¸ â€“ sin(Î¸)), y = a (1 + cos(Î¸))

Solution:

Here, x = a (Î¸ â€“ sin(Î¸)), y = a (1 + cos(Î¸))

= a ()

= a (1 â€“ cos(Î¸))

And, now

= a ()

= a (0 + (- sin (Î¸)))

= – a sin (Î¸)

Now, as

=

(Using identity: sin(2Î¸) = 2 sinÎ¸ cosÎ¸ and 1- cos(2Î¸) = 2 sin2Î¸)

= – cot(Î¸/2)

Solution:

Here, x =

=

And, now

=

Now, as

= – cot 3(t)

### Question 8. x = a (cos(t) + log tan), y = a sin(t)

Solution:

Here, x = a (cos(t) + log tan ), y = a sin(t)

= a ()

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )              (Using identity: 2 sinÎ¸ cosÎ¸ = sin(2Î¸))

= a ( – sin(t))

= a ()

= a ()

=

And, now

= a

= a cos(t)

Now, as

= tan(t)

### Question 9. x = a sec(Î¸), y = b tan(Î¸)

Solution:

Here, x = a sec(Î¸), y = b tan(Î¸)

= a ()

= a (sec(Î¸) tan(Î¸))

= a sec(Î¸) tan(Î¸)

And, now

= b ()

= b (sec2(Î¸))

Now, as

### Question 10. x = a (cos(Î¸) + Î¸ sin(Î¸)), y = a (sin(Î¸) â€“ Î¸ cos(Î¸))

Solution:

Here, x = a (cos(Î¸) + Î¸ sin(Î¸)), y = a (sin(Î¸) â€“ Î¸ cos(Î¸))

= a ()

= a (- sin(Î¸) + (Î¸.) + sin(Î¸).)

= a (- sin(Î¸) + (Î¸.(cos(Î¸) + sin(Î¸).1))

= a (- sin(Î¸) + Î¸ cos(Î¸) + sin(Î¸))

= aÎ¸ cos(Î¸)

And, now

= a ()

= a (cos (Î¸) – (Î¸.) + cos(Î¸).)

= a (cos(Î¸) – (Î¸.(-sin (Î¸) + cos(Î¸).1))

= a (cos(Î¸) + Î¸ sin(Î¸) – cos(Î¸))

= aÎ¸ sin(Î¸)

Now, as

= tan(Î¸)

### Question 11. If x = , y = , show that

Solution:

Here, Let multiply x and y.

xy = (

= ()

= ()                           (Using identity: sin-1Î¸ + cos-1Î¸ = )

Let’s differentiate w.r.t x,

x.+ y. = 0

x. + y = 0

Hence, Proved !!!

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