Inverse Circular Function Exercise: 8.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 8.1

1. Evaluate the following without using the table:

Solution:

a. Sin^-1 1 = $\frac{{\rm{\pi }}}{2}$

b. Sin^-1 = $\left( { - \frac{1}{2}} \right)$ = –sin^-1$\left( {\frac{1}{2}} \right)$ = $ - \frac{{\rm{\pi }}}{6}$

c. Cos^-1$\left( { - \frac{{\sqrt 3 }}{1}} \right)$= π – cos^-1$\left( {\frac{{\sqrt 3 }}{2}} \right)$= π – $\frac{{\rm{\pi }}}{6}$ = $\frac{{5{\rm{\pi }}}}{6}.$

d. Tan^-1(1) = $\frac{{\rm{\pi }}}{4}$

e. Arc cot(-1) = cos^-1(-1) = $\frac{{3{\rm{\pi }}}}{4}$

f. Arc tan$\left( { - \frac{1}{{\sqrt 3 }}} \right)$ = tan^-1$\left( { - \frac{1}{{\sqrt 3 }}} \right)$ = -tan^-1$\left( {\frac{1}{{\sqrt 3 }}} \right)$ = $ - \frac{{\rm{\pi }}}{6}$.

 

2. Express each of the following in terms of x:

Solution:

a. Cos tan^-1x

Let tan^-1x = θ                 [ tan θ = x.]

Now, costan^-1x = cosθ = $\frac{1}{{\sqrt {1 + {{\rm{x}}^2}} }}$ [x = tanθ]

b. sin tan^-1x

Let, cot^-1x = θ                [cotθ = x]

Now, sin.cot^-1x = sinθ = $\frac{1}{{\sqrt {1 + {{\rm{x}}^2}} }}$    [cotθ = x]

c. Tan(Arc cotx) = tancot^-1x = tantan^-1$\frac{1}{{\rm{x}}}$ = $\frac{1}{{\rm{x}}}$.

d. Cos.sin^-1x = cos.cos^-1x $\sqrt {1 - {{\rm{x}}^2}} $ = $\sqrt {1 - {{\rm{x}}^2}} $

e. Tan(2tan^-1x) = tantan^-1$\left( {\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}} \right)$ = $\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}$

f. Sin(2tan^-1x) = sin.sin^-1$\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)$ = $\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}$

g. Cos(2cot^-1x)

Let, cos(2cot^-1x) = cos2θ = $\frac{{{{\cot }^2}\theta  - 1}}{{{{\cot }^2}\theta  + 1}}$ = $\frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}$

h. Cot(2 Arc cotx) = cot(2cot^-1x) = cot cot^-1$\left( {\frac{{{{\rm{x}}^2} - 1}}{{2{\rm{x}}}}} \right)$ = $\frac{{{{\rm{x}}^2} - 1}}{{2{\rm{x}}}}$.

 

3. Evaluate each of the following using the table if necessary:

a. Sin.cos^-1$\left( {\frac{3}{5}} \right)$ = sin.sin^-1$\sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} $ = sin.sin^-1$\frac{4}{5}$ = $\frac{4}{5}$.

b. cos$\left( {{\rm{Arc}}\cos \frac{2}{3}} \right)$ = cos.cos^-1$\frac{2}{3}$ = $\frac{2}{3}$

c. Arc tan $\left( {\tan \frac{{\rm{\pi }}}{6}{\rm{\: }}} \right)$ = tan^-1.tan $\frac{{\rm{\pi }}}{6}$ = $\frac{{\rm{\pi }}}{6}.$

d. Sin $\left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)$ = sin.sin^-1$\frac{3}{5}$ = $\frac{3}{5}$.

e. sin$\left( {2{{\cos }^{ - 1}}\frac{1}{2}} \right)$

Let cos^-1 $\frac{1}{2}$= θ

So, cosθ = $\frac{1}{2}$.

Now, sin$\left( {2{{\cos }^{ - 1}}\frac{1}{2}} \right)$= sin2θ = 2sinθ.cosθ.

= 2. $\frac{1}{2}$

= 1

f. Sin^-1$\left( {2\cos \frac{{\rm{\pi }}}{3}} \right)$ = sin^-1$\left( {\frac{{2.1}}{2}} \right)$ = sin^-1(1) = $\frac{{\rm{\pi }}}{2}$.

 

4.Prove each of the following:

a. 2cos^-1x = cos^-1(2x² – 1)

Solution:

Let, cos^-1x = θ.

So, cosθ = x

Now, cos2θ = 2cos²θ – 1

Or, cos2θ = 2x² – 1   [cosθ = x]

Or, 2θ = cos^-1(2x² – 1)

So, 2cos^-1x = cos^-1(2x² – 1).

 

b. 3cos^-1x = cos^-1(4x³ – 3x)

Solution:

Let, cos^-1x = θ.

So, cosθ = x

Now, cos3θ = 4cos³θ – 3cosθ

Or, cos3θ = 4x³ – 3x   [cosθ = x]

Or, 3θ = cos^-1(4x³ – 3x)

So, 3cos^-1x = cos^-1(4x³ – 3x).

 

c. 3tan^-1x = tan^-1$\frac{{3{\rm{x}} - {{\rm{x}}^3}}}{{1 - 3{{\rm{x}}^2}}}$

Solution:

Let tan^-1x = θ

So, tanθ = x

Now, tan3θ = $\frac{{3\tan \theta  - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}$

Or, tan3θ = $\frac{{3{\rm{x}} - {{\rm{x}}^3}}}{{1 - 3{{\rm{x}}^2}}}$

Or, 3θ = tan^-1$\frac{{3{\rm{x}} - {{\rm{x}}^3}}}{{1 - 3{{\rm{x}}^2}}}$

So, 3tan^-1x = tan^-1$\frac{{3{\rm{x}} - {{\rm{x}}^3}}}{{1 - 3{{\rm{x}}^2}}}$.

 

d. sin(Arc.cost)= cos(Arc.sint)

Solution:

L.H.S. = sin(Arc.cost) = sin(cos^-1) = sin.sin^-1$\sqrt {1 - {{\rm{t}}^2}} $ = $\sqrt {1 - {{\rm{t}}^2}} $.

R.H.S = cos(Arc.sint) = cos.sin^-1t = cos.cos^-1$\sqrt {1 - {{\rm{t}}^2}} $ = $\sqrt {1 - {{\rm{t}}^2}} $.

So, L.H.S. = R.H.S.

 

e. Cos(2 Arc.cost) =2t² – 1

Solution:

Cos(2 Arc.cost) = cos(2cos^-1t) = cos.cos^-1(2t² – 1)

So, cos(2 Arc.cost) = 2t² – 1

 

f. sin(2sin^-1x) = 2x.$\sqrt {1 - {{\rm{x}}^2}} $

Solution:

Let, sin^-1x = θ.

So. sinθ = x.

Now, sin(2sin^-1x) = sin2θ = 2sinθ.cosθ

So, sin(2sin^-1x) = 2x.$\sqrt {1 - {{\rm{x}}^2}} $  [sinθ = x]

 

g. cos(sin^-1u + cos^-1v) = v$\sqrt {1 - {{\rm{u}}^2}} $ – u$\sqrt {1 - {{\rm{v}}^2}} $

Solution:

Let, sin^-1u = A → sinA = u and cosA = $\sqrt {1 - {{\rm{u}}^2}} $

And cos^-1v = b → cosB = v and sinB = $\sqrt {1 - {{\rm{v}}^2}} $

We have,

Cos(A + B) = cosA.cosB – sinA.sinB

So, cos(sin^-1u + cos^-1v) = v$\sqrt {1 - {{\rm{u}}^2}} $ – u$\sqrt {1 - {{\rm{v}}^2}} $

 

h. tan^-1$\frac{1}{5}{\rm{\: }}$+ tan^-1$\frac{1}{7}{\rm{\: }}$= tan^-1$\left( {\frac{6}{{17}}} \right)$                                                                                   

Solution:

L.H.S. = tan^-1$\frac{1}{5}$ + tan^-1$\frac{1}{7}$ = tan^-1$\left( {\frac{1}{5}} \right)$ + tan^-1$\left( {\frac{1}{7}} \right)$ = tan^-1$\frac{{\frac{1}{5} + \frac{1}{7}}}{{1 - \frac{1}{5}.\frac{1}{7}}}$

= tan^-1$\left( {\frac{{7 + 5}}{{35 - 1}}} \right)$ = tan^-1$\left( {\frac{{12}}{{34}}} \right)$

So, tan^-1$\frac{1}{5}{\rm{\: }}$+ tan^-1$\frac{1}{7}{\rm{\: }}$= tan^-1$\left( {\frac{6}{{17}}} \right)$

 

(i) tan^-1 a– tan^-1c= tan^-1$\frac{{{\rm{a}} - {\rm{b}}}}{{1 + {\rm{ab}}}}$ + tan^-1$\frac{{{\rm{b}} - {\rm{c}}}}{{1 + {\rm{bc}}}}$

Solution:

Here, R.H.S. = tan^-1$\frac{{{\rm{a}} - {\rm{b}}}}{{1 + {\rm{ab}}}}$ + tan^-1$\frac{{{\rm{b}} - {\rm{c}}}}{{1 + {\rm{bc}}}}$

= tan^-1a – tan^-1b + tan^-1b – tan^-1c = tan^-1 a– tan^-1c = L.H.S.

 

(j) tan(Arc.tanu – Arc.tanv) = $\frac{{{\rm{u}} - {\rm{v}}}}{{1 + {\rm{uv}}}}$
Solution:

L.H.S. = tan(Arc.tanu – Arc.tanv) = tan(tan^-1u – tan^-1v)

= tan.tan^-1$\frac{{{\rm{u}} - {\rm{v}}}}{{1 + {\rm{uv}}}}$ = $\frac{{{\rm{u}} - {\rm{v}}}}{{1 + {\rm{uv}}}}$ = R.H.S.

 

(k) tan(2tan^-1x) =2 tan (tan^-1x + tan^-1x³)

Solution:

 

(l) tan^-1x + tan^-1y + tan^-1z = tan^-1$\frac{{{\rm{x}} + {\rm{y}} + {\rm{z}} - {\rm{xyz}}}}{{1 - {\rm{xy}} - {\rm{zx}} - {\rm{yz}}}}$

Solution:

L.H.S. = tan^-1x + tan^-1y + tan^-1z.

= tan^-1$\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}} \right)$ + tan^-1z = tan^-1$\frac{{\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}} + {\rm{Z}}} \right)}}{{1 - \frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}.{\rm{z}}}}$

= tan^-1$\frac{{{\rm{x}} + {\rm{y}} + {\rm{z}} - {\rm{xyz}}}}{{1 - {\rm{xy}} - {\rm{zx}} - {\rm{yz}}}}$.

 

(m) sin^-1$\frac{4}{5}$ + sin^-1$\frac{5}{13}$ + sin^-1$\frac{16}{65}$= $\frac{{\rm{\pi }}}{2}$

Solution:

L.H.S. = sin^-1$\left\{ {\frac{4}{5}\sqrt {1 - {{\left( {\frac{5}{{13}}} \right)}^2}}  + \frac{5}{{13}}\sqrt {1 - {{\left( {\frac{4}{5}} \right)}^2}} } \right\}$ + sin^-1$\frac{{16}}{{65}}$.

= sin^-1$\left( {\frac{4}{5}.\frac{{12}}{{13}} + \frac{5}{{13}}.\frac{3}{5}} \right)$ + sin^-1$\frac{{16}}{{65}}$= sin^-1$\left( {\frac{{63}}{{65}}} \right){\rm{\: }}$ + sin^-1$\frac{{16}}{{65}}.$

= sin^-1$\left\{ {\frac{{63}}{{65}}\sqrt {1 - {{\left( {\frac{{16}}{{65}}} \right)}^2}}  + \frac{{16}}{{65}}\sqrt {1 - {{\left( {\frac{{63}}{{65}}} \right)}^2}} } \right\}$

= sin^-1$\left( {\frac{{63}}{{65}}.\frac{{63}}{{65}} + \frac{{16}}{{65}}.\frac{{16}}{{65}}} \right)$ = sin^-1$\frac{{3969 + 256}}{{4225}}$.

= sin^-1$\left( {\frac{{4225}}{{4225}}} \right)$ = sin^-11 = $\frac{{\rm{\pi }}}{2}$ = R.H.S.

 

(n) Prove that: tan^-1$\sqrt {\rm{x}} $ = $\frac{1}{2}$ cos^-1$\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}$ = $\frac{1}{2}$sin^-1$\left( {\frac{{2\sqrt {\rm{x}} }}{{1 + {\rm{x}}}}} \right)$

Solution:

Or, $\frac{1}{2}$cos^-1$\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}$ = $\frac{1}{2}$cos^-1${\rm{\: }}\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$       [x = tan²θ say]

= $\frac{1}{2}$cos^-1 cos2θ = $\frac{1}{2}$.2θ = θ = tan^-1$\sqrt {\rm{x}} $.

And, $\frac{1}{2}$sin^-1$\frac{{2\sqrt {\rm{x}} }}{{1 + {\rm{x}}}}$ = $\frac{1}{2}$ sin^-1$\frac{{2{\rm{tan}}\theta }}{{1 + {{\tan }^2}\theta }}$       [x = tan²θ say]

= $\frac{1}{2}$ sin^-1 sin2θ.

= $\frac{1}{2}$.2θ = θ tan^-1$\sqrt {\rm{x}} {\rm{\: }}$ [x = tan²θ say]

Hence, tan^-1$\sqrt {\rm{x}} $ = $\frac{1}{2}$ cos^-1$\frac{{1 - {\rm{x}}}}{{1 + {\rm{x}}}}$ = $\frac{1}{2}$sin^-1$\left( {\frac{{2\sqrt {\rm{x}} }}{{1 + {\rm{x}}}}} \right)$.

 

5. Find the value of each of the following:

a. cos$\left( {{\rm{si}}{{\rm{n}}^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{5}{{12}}{\rm{\: }}} \right)$

Solution:

Let sin^-1 = $\frac{4}{5}$ = A.

SO, sinA = $\frac{4}{5}$

And, cosA = $\frac{3}{5}$

And tan^-1$\frac{5}{{12}}$ = B

So, tanB = $\frac{5}{{12}}$.

CosB = $\frac{{12}}{{13}}$, sinB = $\frac{5}{{13}}$.

Now, cos$\left( {{\rm{si}}{{\rm{n}}^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{5}{{12}}{\rm{\: }}} \right)$

= cos(A + B) = cosA.cosB – sinA.sinB

= $\frac{3}{5}$.$\frac{{12}}{{13}}$ – $\frac{4}{5}$.$\frac{5}{{13}}$ = $\frac{{36 - 20}}{{65}}$ = $\frac{{16}}{{65}}$.

 

b. sin$\left( {{{\cos }^1}\frac{1}{2} + {{\sin }^{ - 1}}\frac{3}{5}} \right)$

Solution:

Let, cos^-1$\frac{1}{2}$  = A

So, cos A = $\frac{1}{2}$, sinA = $\frac{{\sqrt 3 }}{2}$

And sin^-1$\frac{3}{5}$ = B

So, sinB = $\frac{3}{5}$, cosB = $\frac{4}{5}$.

Now, sin$\left( {{{\cos }^1}\frac{1}{2} + {{\sin }^{ - 1}}\frac{3}{5}} \right)$

= sin(A + B) = sinA.cosB + cosA.sinB

= $\frac{{\sqrt 3 }}{2}$.$\frac{4}{5}$ + $\frac{1}{2}$.$\frac{3}{5}$ = $\frac{{4\sqrt 3  + 3}}{{10}}$.

 

c. tan$\left( {{{\cos }^{ - 1}}\frac{4}{5} - {{\sin }^{ - 1}}\frac{{12}}{{13}}} \right)$

Solution:

Let, cos^-1$\frac{4}{5}$ = A

So, cos A = $\frac{4}{5}$

So, tanA = $\frac{3}{4}$

Or, sin^-1$\frac{{12}}{{13}}$ = B

So, sinB = $\frac{{12}}{{13}}$

So, tanB = $\frac{{12}}{5}$.

Now, tan$\left( {{{\cos }^{ - 1}}\frac{4}{5} - {{\sin }^{ - 1}}\frac{{12}}{{13}}} \right)$ = tan(A – B)

= $\frac{{{\rm{tanA}} - {\rm{tanB}}}}{{1 + {\rm{tanA}}.{\rm{tanB}}}}$ = $\frac{{\frac{3}{4} - \frac{{12}}{5}}}{{1 + \frac{3}{4}.\frac{{12}}{5}}}$ = $\frac{{15 - 48}}{{20 + 36}}$ = $ - \frac{{33}}{{56}}$.

 

d. tan(tan^-1x – tan^-12y)

Solution:

tan(tan^-1x – tan^-12y)

= tan^-1$\left( {\frac{{{\rm{x}} - 2{\rm{y}}}}{{1 + {\rm{x}}.2{\rm{y}}}}} \right)$[Formula of tan^-1x – tan^-1y]

= $\frac{{{\rm{x}} - 2{\rm{y}}}}{{1 + 2{\rm{xy}}}}$

 

e. tan^-13 + tan^-1$\frac{1}{3}$

Solution:

tan^-13 + tan^-1$\frac{1}{3}$

= tan^-1$\frac{{\left( {3 + \frac{1}{3}} \right)}}{{1 - \frac{{3.1}}{3}}}$

= tan^-1$\left( {\frac{{9 + 1}}{{3 - 3}}} \right)$ = tan^-1$\left( {\frac{{10}}{0}} \right)$= tan^-1 ∞ = $\frac{{\rm{\pi }}}{2}$.

 

f. tan^-13 + tan^-1$\frac{1}{3}$

Solution:

Arc sint – Arc cos(-t)

= sin^-1t – cos^-1(-t) = sin^-1t – {π – cos^-1t}

= sin^-1t – π + cos^-1t

= $\frac{{\rm{\pi }}}{2}$ - π  [sin^-1x + cos^-1x = $\frac{{\rm{\pi }}}{2}$]

= $ - \frac{{\rm{\pi }}}{2}$.

 

6. Solve each of the following equations:

a. Cos^-1x - sin^-1x= 0

Solution:

Cos^-1x = sin^-1x

Or, cos^-1x = cos^-1 $\sqrt {1 - {{\rm{x}}^2}} $

Or, x = $\sqrt {1 - {{\rm{x}}^2}} $

Or, x² = 1 – x² [on squaring]

Or. 2x² = 1

So, x =$ \pm $$\frac{1}{{\sqrt 2 }}$.

Since x = $ - \frac{1}{{\sqrt 2 }}$ does not satisfy the given equation

Hence, x = $\frac{1}{{\sqrt 2 }}$.

 

b. Cos^-1x = sin^-1x

Solution:

Here, given equation is:

Tan^-1x – cot^-1x = 0

Or, tan^-1x = tan^-1$\frac{1}{{\rm{x}}}$.

Or, x = $\frac{1}{{\rm{x}}}$

Or, x² = 1

So, x = 1.

 

c. Sin^-1$\frac{{\rm{x}}}{2}$ = cos^-1x

Solution:

Sin^-1$\frac{{\rm{x}}}{2}$ = cos^-1x

Or, sin^-1$\frac{{\rm{x}}}{2}$ = sin^-1$\sqrt {1 - {{\rm{x}}^2}} $

Or, $\frac{{\rm{x}}}{2}$ = $\sqrt {1 - {{\rm{x}}^2}} $

Squaring , we have, $\frac{{{{\rm{x}}^2}}}{4}$ = 1 – x².

Or, $\frac{{{{\rm{x}}^2}}}{4}$ + x² = 1

Or, $\frac{{5{{\rm{x}}^2}}}{4}$ = 1

Or, x² = $\frac{4}{5}$

So, x = $ \pm \frac{2}{{\sqrt 5 }}$

Since x = $ - \frac{2}{{\sqrt 5 }}$ does not satisfy the given equation.

So, x = $\frac{2}{{\sqrt 5 }}$.

 

d.Cos^-1x = cos^-1$\frac{1}{{2{\rm{x}}}}$

Solution:

Her, given equation is:

Cos^-1x = cos^-1$\frac{1}{{2{\rm{x}}}}$.

Or. X = $\frac{1}{{2{\rm{x}}}}$

Or, x² = $\frac{1}{2}$

So, x = $ \pm \frac{1}{{\sqrt 2 }}$.

 

e. tan^-12x=2tan^-1x

Solution:

tan^-12x = tan^-1$\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}$

Or, 2x = $\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}$

Or, 2x – 2x³ = 2x

Or, -2x³ = 0

So, x = 0

 

f. cos^-1x + cos^-12x=$\frac{1}{2}$ π  

Solution:

Here the given equation is, cos^-1x + cos^-12x = $\frac{{\rm{\pi }}}{2}$.

Or, cos^-12x = $\frac{{\rm{\pi }}}{2}$ - cos^-1x.

Or, cos^-12x = sin^-1x

Or,cos^-12x = cos^-1+ $\sqrt {1 - {{\rm{x}}^2}} $

Or, 2x = $\sqrt {1 - {{\rm{x}}^2}} $

Or, 4x² = 1 – x² [on squaring]

Or, 5x² = 1

So, x = $ \pm $$\frac{1}{{\sqrt 5 }}$^.${\rm{\: \: }}$

 

g. Sin^-1x + cos^-1(1 – x) = $\frac{{\rm{\pi }}}{2}{\rm{\: }}$

Solution:

Sin^-1x + cos^-1(1 – x) = $\frac{{\rm{\pi }}}{2}{\rm{\: }}$

Or, cos^-1(1 – x) = $\frac{{\rm{\pi }}}{2}$ - sin^-1x

Or, cos^-1(1 – x) = cos^-1 x

Or, 1 – x = x

Or, 1 = 2x

So, x = $\frac{1}{2}$.

 

h. Sin^-1 $\frac{{2{\rm{a}}}}{{1 + {{\rm{a}}^2}}}{\rm{\: }}$- cos^-1$\frac{{1 - {{\rm{b}}^2}}}{{1 + {{\rm{b}}^2}}}$ = 2tan^-1x

Solution:

The given equation is:

Sin^-1 $\frac{{2{\rm{a}}}}{{1 + {{\rm{a}}^2}}}{\rm{\: }}$- cos^-1$\frac{{1 - {{\rm{b}}^2}}}{{1 + {{\rm{b}}^2}}}$ = 2tan^-1x.

Or, 2tan^-1a – 2tan^-1b = 2tan^-1x  [2tan^-1x = sin^-1$\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}$.]

Or, tan^-1a – tan^-1b = tan^-1x

Or, tan^-1$\left( {\frac{{{\rm{a}} - {\rm{b}}}}{{1 + {\rm{ab}}}}} \right)$ = tan^-1x

Or, x = $\frac{{{\rm{a}} - {\rm{b}}}}{{1 + {\rm{ab}}}}$

So, x = $\frac{{{\rm{a}} - {\rm{b}}}}{{1 + {\rm{ab}}}}$.

 

i. tan^-1$\frac{{{\rm{x}} - 1}}{{{\rm{x}} - 2}}$ + tan^-1$\frac{{{\rm{x}} + 1}}{{{\rm{x}} + 2}}$ = tan^-11

Solution:

Given equation is tan^-1$\frac{{{\rm{x}} - 1}}{{{\rm{x}} - 2}}$ + tan^-1$\frac{{{\rm{x}} + 1}}{{{\rm{x}} + 2}}$= tan^-11.

Or, tan^-1$\left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{x}} - 1}}{{{\rm{x}} - 2}} + \frac{{{\rm{x}} + 1}}{{{\rm{x}} + 2}}}\\{1 - \frac{{{\rm{x}} - 1}}{{{\rm{x}} - 2}}.\frac{{{\rm{x}} + 1}}{{{\rm{x}} + 2{\rm{\: }}}}}\end{array}} \right\}$ = tan^-1 1.

Or, $\frac{{\left( {{\rm{x}} + 1} \right)\left( {{\rm{x}} + 2} \right) + \left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} + 1} \right)}}{{\left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} + 1} \right) - \left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right)}} = 1$

Or, $\frac{{{{\rm{x}}^2} + 2{\rm{x}} - {\rm{x}} - 2 + {{\rm{x}}^2} + {\rm{x}} - 2{\rm{x}} - 2}}{{{{\rm{x}}^2} - 4 - \left( {{{\rm{x}}^2} - 1} \right)}}$ = 1

0r, 2x² – 4 = -3

Or, 2x² = 1

Or, x² = $\frac{1}{2}$

So, x = $ \pm \frac{1}{{\sqrt 2 }}$.

 

j.tan^-12x + tan^-13x = $\frac{{\rm{\pi }}}{4}$

Solution:

The given equation is:

Tan^-12x + tan^-13x = $\frac{{\rm{\pi }}}{4}$.

Or, tan^‑1$\left( {\frac{{2{\rm{x}} + 3{\rm{x}}}}{{1 - 2{\rm{x}}.3{\rm{x}}}}} \right)$ ⁼ $\frac{{\rm{\pi }}}{4}$^.

Or, $\frac{{5{\rm{x}}}}{{1 - 6{{\rm{x}}^2}}}$ = tan$\frac{{\rm{\pi }}}{4}$ = 1

Or, 5x – 1 – 6x² = 0

Or, 6x² + 5x – 1 = 0

Or, 6x² + 6x – x – 1 = 0

Or, 6x(x + 1) – 1 (x + 1) = 0

Or, (x + 1)(6x – 1) = 0

So, x = -1, $\frac{1}{6}$.

 

k. 3tan^-1$\frac{1}{{2 + \sqrt 3 }}$ - tan^-1$\frac{1}{{\rm{x}}}$ = tan^-1$\frac{1}{3}$

Solution:  The given equation is:

3tan^-1$\frac{1}{{2 + \sqrt 3 }}$ - tan^-1$\frac{1}{{\rm{x}}}$ = tan^-1$\frac{1}{3}$.

Or, 3tan^-1 (2 – $\sqrt 3 $) – tan^-1$\frac{1}{{\rm{x}}}$ = tan^-1$\frac{1}{3}.$                [3tan^-1x = tan^-1$\left( {\frac{{3{\rm{x}} - {{\rm{x}}^3}}}{{1 - 3{{\rm{x}}^2}}}} \right)$]

Or, tan^-1$\left\{ {\frac{{\left( {3\left( {2 - \sqrt 3 } \right) - {{\left( {2 - \sqrt 3 } \right)}^2}} \right)}}{{1 - 3{{\left( {2 - \sqrt 3 } \right)}^2}}}} \right\}$ – tan^-1$\frac{1}{3}{\rm{\: }}$= tan^-1$\frac{1}{{\rm{x}}}$

Or, tan^-1$\left\{ {\frac{{12\sqrt 3  - 20}}{{12\sqrt 3  - 20}}} \right\}$ - tan^-1$\frac{1}{3}$ = tan^-1$\frac{1}{{\rm{x}}}$.

Or, tan^-1$\left( {\frac{{1 - \frac{1}{3}}}{{1 + \frac{{1.1}}{3}}}} \right)$ = tan^-1$\frac{1}{{\rm{x}}}$

Or, tan^-1$\left( {\frac{2}{3}.\frac{3}{4}} \right)$ = tan^-1$\frac{1}{{\rm{x}}}$.

Or, tan^-1$\frac{1}{2}$ = tan^-1$\frac{1}{{\rm{x}}}$.

Or, $\frac{1}{2}$ = $\frac{1}{{\rm{x}}}$

So, x = 2.

 

7. Prove that:

a. x+y+z=xyz, if tan^-1x+ tan^-1y+tan^-1z = π

Solution:

Tan^-1 + tan^-1y + tan^-1z = π

Or, tan^-1$\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}} \right)$ = π – tan^-1z

Or, $\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}$ = tan (π – tan^-1z)

Or, $\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}$ = -z   [tan(π – tan^-1) = –tan.tan^-1z = -z]

Or, x + y = -z + xuz

So, x + y + z = xyz.

 

b. xy + yz + zx =1, if Tan^-1 + tan^-1y + tan^-1z = $\frac{{\rm{\pi }}}{2}$.

Solution:

Tan^-1 + tan^-1y + tan^-1z = $\frac{{\rm{\pi }}}{2}$

Or, tan^-1$\left( {\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}} \right)$ = $\frac{{\rm{\pi }}}{2}$ – tan^-1z

Or, $\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}$ = tan ($\frac{{\rm{\pi }}}{2}$ – tan^-1z) = cot.tan^-1z.

Or, $\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}$ = -z   [tan(π – tan^-1) = -tan.tan^-1z = -z]

Or, $\frac{{{\rm{x}} + {\rm{y}}}}{{1 - {\rm{xy}}}}$= cot.cot^-1$\frac{1}{{\rm{z}}}$ = $\frac{1}{{\rm{z}}}$

Or, zx + yz = 1 – xy

So, xy + yz + zx = 1.

 

8. If cos^-1x + cos^-1y + cos^-1z = π, show that x² + y² + z² + 2xyz = 1.

Solution:

Here, cos^-1x + cos^-1y + cos^-1z = π

Or, cos^-1 { xy - $\sqrt {\left( {1 - {{\rm{x}}^2}} \right)\left( {1 - {{\rm{y}}^2}} \right)} $ } = π – cos^-1z

Or, xy - $\sqrt {\left( {1 - {{\rm{x}}^2}} \right)\left( {1 - {{\rm{y}}^2}} \right)} $ = cos (π – cos^-1z)

Or, xy - $\sqrt {\left( {1 - {{\rm{x}}^2}} \right)\left( {1 - {{\rm{y}}^2}} \right)} $ = -cos.cos^-1z = -z

Or, xy + z = $\sqrt {\left( {1 - {{\rm{x}}^2}} \right)\left( {1 - {{\rm{y}}^2}} \right)} $

Squaring, we have,

X²y² + 2xyz + z² = 1 – y² – x² + x²y².

So, x² + y² + z² + 2xyz = 1.

 

9. Prove that:

(i) tan^-1$\frac{3}{5}$ + sin^-1$\frac{3}{5}$ = tan^-1$\frac{{27}}{{11}}$

Solution:

Let, sin^-1$\frac{3}{5}$ = θ →sin θ = $\frac{3}{5}$.

So, p = 3 , b = 4, h = 5.

So,tanθ = $\frac{3}{4}$.

Or, θ = tan^-1$\frac{3}{4}$.

So, sin^-1$\frac{3}{5}$ = tan^-1$\frac{3}{4}$.

L.H.S. = tan^-1$\frac{3}{5}$ + sin^-1$\frac{3}{5}$ = tan^-1$\frac{3}{5}$ + tan^-1$\frac{3}{4}$.

= tan^-1$\frac{{\frac{3}{5} + \frac{3}{4}}}{{1 - \frac{3}{5}.\frac{3}{4}}}$ = tan^-1$\frac{{12 + 15}}{{20 - 9}}$ = tan^-1$\frac{{27}}{{11}}$ = R.H.S.

 

(ii) tan^-1$\frac{1}{3}$+ tan^-1$\frac{1}{5}$ + tan^-1$\frac{1}{7}$ + tan^-1$\frac{1}{8}$=$\frac{{\rm{\pi }}}{4}$

Solution:

L.H.S. = tan^-1$\frac{1}{3}$+ tan^-1$\frac{1}{5}$ + tan^-1$\frac{1}{7}$ + tan^-1$\frac{1}{8}$.

= tan^-1$\frac{{\frac{1}{3} + \frac{1}{5}}}{{1 - \frac{1}{3}.\frac{1}{5}}}$ + tan^-1$\frac{{\frac{1}{7} + \frac{1}{8}}}{{1 - \frac{1}{7}.\frac{1}{8}}}$

= tan^-1$\left( {\frac{{5 + 3}}{{14}}} \right)$ + tan^-1$\left( {\frac{{8 + 7}}{{55}}} \right)$ = tan^-1$\frac{4}{7}$ + tan^-1$\frac{3}{{11}}$.

= tan^-1$\frac{{\frac{4}{7} + \frac{3}{{11}}}}{{1 - \frac{4}{7}.\frac{3}{{11}}}}$ = tan^-1$\frac{{44 + 21}}{{77 - 12}}$ = tan^-1$\left( {\frac{{65}}{{65}}} \right)$ = tan^-1 1

= $\frac{{\rm{\pi }}}{4}$ = R.H.S.

 

(iii) 2tan^-1$\frac{1}{3}$ + tan^-1$\frac{1}{7}$ = $\frac{{\rm{\pi }}}{4}$

L.H.S. = 2tan^-1$\frac{1}{3}$ + tan^-1$\frac{1}{7}$

= tan^-1$\left( {\frac{{2.\frac{1}{3}}}{{1 - \frac{1}{9}{\rm{\: }}}}} \right)$ + tan^-1$\frac{1}{7}$   [2tan^-1x = tan^-1$\left( {\frac{{2{\rm{x}}}}{{1 - {{\rm{x}}^2}}}} \right)$]

= tan^-1$\left( {\frac{2}{3}.\frac{9}{8}} \right)$ + tan^-1$\frac{1}{7}$ = tan^-1$\frac{3}{4}$ + tan^-1$\frac{1}{7}$.

= tan^-1$\left( {\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 - \frac{3}{4}.\frac{1}{7}}}} \right)$ = tan^-1$\left( {\frac{{21 + 4}}{{28 - 3}}} \right)$ = tan^-1$\left( {\frac{{25}}{{25}}} \right)$

= tan^-1 1 = $\frac{{\rm{\pi }}}{4}$ = R.H.S.

 

(iv) 4(cot^-13 + cot^-1 2) = π

Solution:

Cosec^-1$\sqrt 5 $= θ.

So, cosec θ = $\sqrt 5 $,  (p = 1, h = $\sqrt 5 $, b = 2)

So, cot θ = 2

So, θ = cot^-1 2

So, cosec^-1$\sqrt 5 $ = cos^-12 …(i)

L.H.S. = 4(cot^-1 3 + cosec^-1$\sqrt 5 $)

= 4(cot^-13 + cot^-1 2)    [from(i)]

= 4 $\left( {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2}} \right)$ = 4tan^-1$\left( {\frac{{\frac{1}{3} + \frac{1}{2}}}{{1 - \frac{1}{3}.\frac{1}{2}}}} \right)$ = 4tan^-1$\left( {\frac{5}{5}} \right)$

= 4tan^-1 1 = 4. $\frac{{\rm{\pi }}}{4}$ = π = R.H.S.

 

(v) tan^-11 + tan^-12 + tan^-13 = π

Solution:

L.H.S. = tan^-11 + tan^-12 + tan^-13

= tan^-1 + tan^-1$\left( {\frac{{2 + 3}}{{1 - 2.3}}} \right)$ = tan^-1.tan^-1(-1) = $\frac{{\rm{\pi }}}{4}$ + $\frac{{3{\rm{\pi }}}}{4}$ = π.

R.H.S. = 2$\left( {{\rm{ta}}{{\rm{n}}^{ - 1}}1 + {{\tan }^{ - 1}}\frac{1}{2} + {{\tan }^{ - 1}}\frac{1}{3}} \right)$

= 2$\left\{ {{{\tan }^{ - 1}}1 + {{\tan }^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2}.\frac{1}{3}}}} \right){\rm{\: }}} \right\}$

= 2$\left\{ {{{\tan }^{ - 1}}1 + {{\tan }^{ - 1}}\left( {\frac{5}{6}.\frac{6}{4}} \right)} \right\}$

= 2(tan^-11 + tan^-11) = 2$\left( {\frac{{\rm{\pi }}}{4} + \frac{{\rm{\pi }}}{4}} \right)$ = π.

  

10. If sin^-1x + Sin^-1y + sin^-1z = π, prove that x$\sqrt {1 - {{\rm{x}}^2}} $ + y$\sqrt {1 - {{\rm{y}}^2}} $ + z$\sqrt {1 - {{\rm{z}}^2}} $.

Solution:

Let, sin^-1x = A → sinA = x and cosA = $\sqrt {1 - {{\rm{x}}^2}} $

Sin^-1y = B → sinB = y and cosB = $\sqrt {1 - {{\rm{y}}^2}} $

And sin^-1z = C → sinC = z and cosC = $\sqrt {1 - {{\rm{z}}^2}} $

And sin^-1x + sin^-1 y + sin^-1 z = π.

Or, A + B + C = π.

So, A + B = π – c

Or, sin(A + B) = sin(π – C) = sinC.

Cos(A + B) = cos(π – C) = - cosC

Now,

L.H.S. = x$\sqrt {1 - {{\rm{x}}^2}} $ + y$\sqrt {1 - {{\rm{y}}^2}} $ + z$\sqrt {1 - {{\rm{z}}^2}} $

= sinA.cosA + sinB.cosB + sinC.cosC.

= $\frac{1}{2}$ (2sinA.cosA +2 sinB.cosB) + sinC.cosC

= $\frac{1}{2}$ (sin2A + sin2B) + sinC.cosC

= $\frac{1}{2}$.2 sin$\frac{{2{\rm{A}} + 2{\rm{B}}}}{2}$ .cos$\frac{{2{\rm{A}} - 2{\rm{B}}}}{2}$ + sinC.cosC.

= sin(A + B). cos(A – B) + sinC.cosC.

= sinC.cos(A – B) + sinC.cosC                   [A + B = π – C]

= sinC {cos(A – B) + cosC}

= sinC {cos(A – B) – cos(A + B)}               [A + B = π – C]

= sinC.2sinA.sinB = 2sinA.sinB.sinC = 2xy = R.H.S.

Hence, x$\sqrt {1 - {{\rm{x}}^2}} $ + y$\sqrt {1 - {{\rm{y}}^2}} $ + z$\sqrt {1 - {{\rm{z}}^2}} $

 

11. If sin^-1x + Sin^-1y + sin^-1z = $\frac{π}{2}$, prove that x² + y² + z² + 2xyz = 1.

Solution:

Proof:

Let sin^-1x → x = sinA

Sin^-1y = B → y sinB

And sin^-1z = C → z = sinC.

Then the given question changed to:

If  A + B + C = $\frac{{\rm{\pi }}}{2}$, prove that,

Sin²A + sin²B + sin²C = 1 – 2sinA.sinB.sinC.

From the given part, A + B = $\frac{{\rm{\pi }}}{2}$ – C.

Sin(A + B) = cosC

And cos (A + B) = sinC.

Now, L.H.S.= $\frac{1}{2}$(2sin²A + 2sin²B) + sin²C.

= $\frac{1}{2}$ [(1 – cos2A) + (1 – cos2B)] + sin²C.

= $\frac{1}{2}$ [2 – cos2A – cos2B] + sin²c.

= 1 – $\frac{1}{2}$ (cos2A + cos2B) + sin²C.

= 1 – $\frac{1}{2}$ 2cos(A + B).cos(A – B) + sin²C.

= 1 – sinC [cos(A – B) – cos(A + B)]

= 1 – sinC $\left[ {2\sin \frac{{{\rm{A}} - {\rm{B}} + {\rm{A}} + {\rm{B}}}}{2}.\sin \frac{{{\rm{A}} + {\rm{B}} - {\rm{A}} + {\rm{B}}}}{2}} \right]$

= 1 – sinC [2sinA.sinB]

= 1 – 2sinA.sinB.sinC = R.H.S>

Hence, x² + y² + z² + 2xyz = 1.