(a,c) Let x ⇒ x + h and y → x

|tan –1 x – tan –1 y| ≤ |x – y|

|tan –1 (x + h) – tan –1 x| ≤ |h|

≤ 1

≤ 1 hence true

|sin x – sin y| ≤ |x – y|

x → x + h y → x

≤ 1

|cos x| ≤ 1 hence true

Alternative solutions

For x = y, this is true

∴ Let x, y ∈ R and x < y

consider f(t) = tan –1 t, t ∈ [x, y]

Using LMVT, = , c ∈ (x, y)

⇒ tan –1 y – tan –1 x = ≤ y – x ........(i)

similarly x > y, tan –1 x – tan –1 y ≤ x – y ........(ii)

From (i) and (ii) we get

≤ |x – y|

Similarly considering g(t) = sin t in [x, y]

we get = cos c

⇒ sin y – sin x = (cos c) (y – x) ≤ y – x ........(iii)

and sin x – sin y ≤ x – y ........(iv)

(iii), (iv) ⇒ |sin x – siny| ≤ |x – y|