Let f(x) = (sin(tan--1x) + sin(cot--1x))2 -- 1, |x| 1.

Question

Let f(x) = (sin(tan--1x) + sin(cot--1x))2 -- 1, |x| > 1.

If $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$ and $$y\left( {\sqrt 3 } \right) = {\pi  \over 6}$$,
then y($${ - \sqrt 3 }$$) is equal to :

Accepted Answer

Given f(x) = (sin(tan--1x) + sin(cot--1x))2 -- 1

= (sin(tan--1x) + sin($${\pi  \over 2}$$ - tan--1x))2 -- 1

= (sin(tan--1x) + cos(tan--1x))2 -- 1

= sin2(tan--1x) + cos2(tan--1x) + 2sin(tan--1x)cos(tan--1x) + 1

= 1 + sin(2tan--1x) - 1

= sin(2tan--1x)

Also given $${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$$

Integrating both sides we get

y = $${1 \over 2}$$ sin-1 (f(x)) + C

= $${1 \over 2}$$ sin-1 (sin(2tan--1x)) + C

Given $$y\left( {\sqrt 3 } \right) = {\pi  \over 6}$$ mean x = $$\sqrt 3 $$ and y = $${\pi  \over 6}$$

$$ \therefore $$ $${\pi  \over 6}$$ = $${1 \over 2}$$ sin-1 (sin(2tan--1$$\sqrt 3 $$)) + C

$$ \Rightarrow $$ $${\pi  \over 6}$$ = $${1 \over 2}$$ sin-1 (sin(2$$ \times $$ ${\pi  \over 3}$)) + C$ \Rightarrow $${\pi  \over 6}$=${1 \over 2}$sin-1 (${{\sqrt 3 } \over 2}$) + C$ \Rightarrow $${\pi  \over 6}$=${1 \over 2}$$ \times $${\pi  \over 3}$+ C$ \Rightarrow $C = 0

Now y(${ - \sqrt 3 }$) means when x =${ - \sqrt 3 }$then find y.

y =${1 \over 2}$sin-1 (sin(2tan--1x))

=${1 \over 2}$sin-1 (sin(2tan--1(${ - \sqrt 3 }$)))

=${1 \over 2}$sin-1 (sin(-2tan--1(${  \sqrt 3 }$)))

=${1 \over 2}$sin-1 (sin(-2$ \times $$ ${\pi  \over 3}$$))

= $${1 \over 2}$$ sin-1 (-sin(2$$ \times $$ ${\pi  \over 3}$))

=${1 \over 2}$sin-1 (-${{\sqrt 3 } \over 2}$)

=${1 \over 2}$$ \times $-sin-1 (${{\sqrt 3 } \over 2}$)

=${1 \over 2}$$ \times $-${\pi  \over 3}$= -${\pi  \over 6}$

Solutions · 1