Inverse Trigonometric Functions

The star "*" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem.

In this case, the operation "*" is defined by the equation $x * y = x^2 + y^3$, which means if you have two numbers $x$ and $y$, then the result of applying the "*" operation to them is $x^2 + y^3$.

The problem also specifies an additional rule for this operation: $(x * 1) * 1 = x * (1 * 1)$, which needs to be taken into account when solving the problem. This is a type of "associativity" condition.

Given,

$x\, * \,y = {x^2} + {y^3}$

$\therefore$ $x\, * \,1 = {x^2} + {1^3} = {x^2} + 1$

Now, $(x\, * \,1)\, * \,1 = ({x^2} + 1)\, * \,1$

$ \Rightarrow (x\, * \,1)\, * \,1 = {({x^2} + 1)^2} + {1^3}$

$ \Rightarrow (x\, * \,1)\, * \,1 = {x^4} + 1 + 2{x^2} + 1$

Also, $x\, * \,(1\, * \,1)$

$ = x\, * \,({1^2} + {1^3})$

$ = x\, * \,2$

$ = {x^2} + {2^3}$

$ = {x^2} + 8$

Given that,

$(x\, * \,1)\, * \,1 = x\, * \,(1\, * \,1)$

$\therefore$ ${x^4} + 1 + 2{x^2} + 1 = {x^2} + 8$

$ \Rightarrow {x^4} + {x^2} - 6 = 0$

$ \Rightarrow {x^4} + 3{x^2} - 2{x^2} - 6 = 0$

$ \Rightarrow {x^2}({x^2} + 3) - 2({x^3} + 3) = 0$

$ \Rightarrow ({x^2} + 3)({x^2} - 2) = 0$

$ \Rightarrow {x^2} = 2,\, - 3$

[${x^2} = -3$ not possible as square of anything should be always positive]

$\therefore$ ${x^2} = 2$

$\therefore$ Now,

$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{{{2^2} + 2 - 2} \over {{2^2} + 2 + 2}}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{4 \over 8}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{1 \over 2}} \right)$

$ = 2 \times {\pi \over 6}$

$ = {\pi \over 3}$

${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$

Let $f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$

Where $t = {\tan ^{ - 1}}x$ ; $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$

$ = {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3\pi } \over 2}{t^2} - {t^3}$

$f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}$

This is a quadratic equation of t.

Here, coefficient of t² term is ${{3\pi } \over 2}$ which is > 0.

$\therefore$ It is a upward parabola.

Now, $f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}$

$f''(t) = 3\pi > 0$

$\therefore$ $3\pi t - {{3{\pi ^2}} \over 4} = 0$

$ \Rightarrow t = {\pi \over 4}$ (minima)

$\therefore$ vertex of graph at ${\pi \over 4}$

$\therefore$ Minimum value at ${\pi \over 4}$ and maximum value at $-$${\pi \over 2}$.

$\therefore$ $f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$

$f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$

$ = {{7{\pi ^3}} \over 8}$

$\therefore$ $k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$

$ \Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$

$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$

$ \frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0 $

The solution to this inequality is

$x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$

for $x^{2}-3 x+2>0$ and $\neq 1$

$ x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\} $

Combining the two solution sets (taking intersection)

$ x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\} $

${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$

$ = (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )$

$ = 4\pi - 11$.

$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$

$ - 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty $ .... (1)

$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left[ {{{ - 1} \over 4},{1 \over 2}} \right] \cup \{ 0\} $ .... (2)

(1) & (2)

$\Rightarrow$ Domain = $\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $

Let g(x) = sin x + cos x = $\sqrt 2 $ sin$\left( {x + {\pi \over 4}} \right)$

g(x)$\in$ $\left[ {1,\sqrt 2 } \right]$ for x$\in$ [0, $\pi$/2]

f(x) = tan^$-$1 (sin x + cos x) $\in$ $\left[ {{\pi \over 4},{{\tan }^{ - 1}}\sqrt 2 } \right]$

tan$({\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}) = {{\sqrt 2 - 1} \over {1 + \sqrt 2 }} \times {{\sqrt 2 - 1} \over {\sqrt 2 - 1}} = 3 - 2\sqrt 2 $

Given $a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$

$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$

$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$

$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$

$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$

$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$

${{1 + x} \over x} \in ( - \infty , - 1] \cup [1,\infty )$

${1 \over x} \in ( - \infty , - 2] \cup [0,\infty )$

$x \in \left[ { - {1 \over 2},0} \right) \cup (0,\infty )$

$x \in \left[ { - {1 \over 2},0} \right) \cup \{ 0\} $

$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} } $

= $\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $

= ${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$

$O \le {x^2} - x + 1 \le 1$

$ \Rightarrow {x^2} - x \le 0$

$ \Rightarrow x \in [0,1]$

Also, $0 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2}$

$ \Rightarrow 0 < {{2x - 1} \over 2} \le 1$

$ \Rightarrow 0 < 2x - 1 \le 2$

$1 < 2x \le 3$

${1 \over 2} < x \le {3 \over 2}$

Taking intersection

$x \in \left( {{1 \over 2},1} \right]$

$ \Rightarrow \alpha = {1 \over 2},\beta = 1$

$ \Rightarrow \alpha + \beta = {3 \over 2}$

$2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) = {\tan ^{ - 1}}\left( {{{6/5} \over {1 - {9 \over {{2^5}}}}}} \right) = {\tan ^{ - 1}}\left( {{{{6 \over 5}} \over {{{16} \over {25}}}}} \right) = {\tan ^{ - 1}}{{15} \over 8}$

$\therefore$ $2{\tan ^{ - 1}}\left( {{3 \over 5}} \right) + {\sin ^{ - 1}}\left( {{5 \over {13}}} \right) = {\tan ^{ - 1}}\left( {{{15} \over 8}} \right) + {\tan ^{ - 1}}\left( {{5 \over {12}}} \right)$

$ = {\tan ^{ - 1}}\left( {{{{{15} \over 8} + {5 \over {12}}} \over {1 - {{15} \over 8},{5 \over {12}}}}} \right)$

$ = {\tan ^{ - 1}}\left( {{{180 + 40} \over {21}}} \right) = {\tan ^{ - 1}}\left( {{{220} \over {21}}} \right)$

${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$

For equation to be defined,

x² + x $\ge$ 0

$\Rightarrow$ x² + x + 1 $\ge$ 1

$\therefore$ Only possibility that the equation is defined

x² + x = 0 $\Rightarrow$ x = 0; x = $-$1

None of these values satisfy

$\therefore$ No of roots = 0

There are three cases possible for $x \in [ - 1,1]$

Case I : $x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$

$ \therefore $ ${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$

$ \Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi $

$ \Rightarrow x = \pm \sqrt \pi $ $ \to $ (Reject)

Case II : $x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)$

$ \therefore $ ${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}$

$ \Rightarrow 0 + \pi = {x^2}$

$ \Rightarrow x = \pm \sqrt x $ $ \to $ (Reject)

Case III : $x \in \left( {\sqrt {{2 \over 3}} ,1} \right)$

$ \therefore $ ${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}$

$ \Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x $ (Reject)

$ \therefore $ No solution. There, the correct answer is (1).

tan^$-$1(x + 1) + cot^$-$1$\left( {{1 \over {x - 1}}} \right)$ = tan^$-$1$\left( {{8 \over {31}}} \right)$

$ \Rightarrow $ tan^$-$1(x + 1) + tan^$-$1(x - 1) = tan^$-$1$\left( {{8 \over {31}}} \right)$

$ \Rightarrow $ ${\tan ^{ - 1}}\left( {{{\left( {x + 1} \right) + \left( {x - 1} \right)} \over {1 - \left( {x + 1} \right)\left( {x - 1} \right)}}} \right)$ = tan^$-$1$\left( {{8 \over {31}}} \right)$

$ \Rightarrow $ ${{(1 + x) + (x - 1)} \over {1 - (1 + x)(x - 1)}} = {8 \over {31}}$

$ \Rightarrow {{2x} \over {2 - {x^2}}} = {8 \over {31}}$

$ \Rightarrow 4{x^2} + 31x - 8 = 0$

$ \Rightarrow x = - 8,{1 \over 4}$

but at $x = {1 \over 4}$

$LHS > {\pi \over 2}$ and $RHS < {\pi \over 2}$

So, only solution is x = $-$ 8 = $-$${{{32} \over 4}}$

${\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100$ terms

$ = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100$ term

$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}} $

$ = \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} } $

$ = \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)} $

$ = {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1$

$ = {\tan ^{ - 1}}{{200} \over {202}}$

$ = {\cot ^{ - 1}}(1.01)$

Hence $\alpha = 1.01$

${\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x$

${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$

${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x$

$x = 0 $ or $3\sqrt {25 - 16{x^2}} + 4\sqrt {25 - 9{x^2}} = 25$

$4\sqrt {25 - 9{x^2}} = 25 - 3\sqrt {25 - 16{x^2}} $

Squaring we get

$16(25 - 9{x^2}) = 625 - 9(25 - 16{x^2}) - 150\sqrt {25 - 16{x^2}} $

$400 = 625 + 225 - 150\sqrt {25 - 16{x^2}} $

$\sqrt {25 - 16{x^2}} = 3 \Rightarrow 25 - 16{x^2} = 9$

$ \Rightarrow {x^2} = 1$

Put x = 0, 1, $-$1 in the original equation

We see that all values satisfy the original equation.

Number of solution = 3

tan^$-$1a + tan^$-$1b = ${\pi \over 4}$ 0 < a, b < 1

$ \Rightarrow {{a + b} \over {1 - ab}} = 1$

a + b = 1 $-$ ab

(a + 1)(b + 1) = 2

Now $\left[ {a - {{{a^2}} \over 2} + {{{a^3}} \over 3} + ....} \right] + \left[ {b - {{{b^2}} \over 2} + {{{b^3}} \over 3} + ....} \right]$

$ = {\log _e}(1 + a) + {\log _e}(1 + b)$

($ \because $ expansion of log_e(1 + x))

$ = {\log _e}[(1 + a)(1 + b)]$

$ = {\log _e}2$

${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$

${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$

Now, ${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}$

$2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}$

$ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}$

$\cos ec\left( {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$

$\cos ec\left( {2{{\tan }^{ - 1}}\left( {{1 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$

$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{{2\left( {{1 \over 5}} \right)} \over {1 - {{\left( {{1 \over 5}} \right)}^2}}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$

$ = \cos ec\left( {{{\tan }^{ - 1}}\left( {{5 \over {12}}} \right) + {{\cos }^{ - 1}}\left( {{4 \over 5}} \right)} \right)$

Let ${\tan ^{ - 1}}(5/12) = \theta \Rightarrow \sin \theta = {5 \over {13}},\cos \theta = {{12} \over {13}}$

and ${\cos ^{ - 1}}\left( {{4 \over 5}} \right) = \phi \Rightarrow \cos \phi = {4 \over 5}$ and $\sin \phi = {3 \over 5}$

$ = \cos ec(\theta + \phi )$

$ = {1 \over {\sin \theta \cos \phi + \cos \theta \sin \phi }}$

$ = {1 \over {{5 \over {13}}.{4 \over 5} + {{12} \over {13}}.{3 \over 5}}} = {{65} \over {56}}$

$\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$

${\sin ^{ - 1}}\left( {{{\sqrt {63} } \over 8}} \right) = \theta $ $\sin \theta = {{\sqrt {63} } \over 8}$



$\cos \theta = {1 \over 8}$

$2{\cos ^2}{\theta \over 2} - 1 = {1 \over 8}$

${\cos ^2}{\theta \over 2} = {9 \over {16}}$

$\cos {\theta \over 2} = {3 \over 4}$

${{1 - {{\tan }^2}{\theta \over 4}} \over {1 + {{\tan }^2}{\theta \over 4}}} = {3 \over 4}$

$\tan {\theta \over 4} = {1 \over {\sqrt 7 }}$

S = ${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$

= ${\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}} \right) + ...$

$ \therefore $ T_r = ${\tan ^{ - 1}}\left( {{1 \over {1 + r \times \left( {r + 1} \right)}}} \right)$

= tan^–1(r + 1) – tan^–1r

$ \therefore $ T₁ = tan^–12 – tan^–11

T₂ = tan^–13 – tan^–12

T₃ = tan^–14 – tan^–13
.
.
.

T₁₀ = tan^-111 – tan^–110

$ \therefore $ S = tan^–111 – tan^–11 = ${\tan ^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)$

$ \therefore $ tan(S) = $\tan \left( {{{\tan }^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)} \right)$

= ${{{11 - 1} \over {1 + 11}}}$ = ${{10} \over {12}} = {5 \over 6}$

$2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$

$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$

$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}$

$ = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$

= $2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)$

$ = 2\pi - {\pi \over 2}$

$ = {{3\pi } \over 2}$

f(x) = ${\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)$

$ \therefore $ $ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}} \le 1$

Since |x| + 5 & x² + 1 is always positive

So ${{\left| x \right| + 5} \over {{x^2} + 1}} \ge 0$

That means this inequality $ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}}$ always right. So we can ignore it.

So for domain :

${{\left| x \right| + 5} \over {{x^2} + 1}} \le 1$

$ \Rightarrow $ ${{x^2} - \left| x \right| - 4 \ge 0}$

$ \Rightarrow $ $\left( {\left| x \right| - {{1 - \sqrt {17} } \over 2}} \right)\left( {\left| x \right| - {{1 + \sqrt {17} } \over 2}} \right) \ge 0$

$ \Rightarrow $ |x| $ \ge $ ${{{1 + \sqrt {17} } \over 2}}$ or |x| $ \le $ ${{{1 - \sqrt {17} } \over 2}}$

As ${{{1 - \sqrt {17} } \over 2}}$ is < 0 and |x| always $ \ge $ 0. So
|x| $ \le $ ${{{1 - \sqrt {17} } \over 2}}$ not possible.

$ \therefore $ |x| $ \ge $ ${{{1 + \sqrt {17} } \over 2}}$

x $ \in $ $\left( { - \infty , - {{1 + \sqrt {17} } \over 2}} \right) \cup \left( {{{1 + \sqrt {17} } \over 2},\infty } \right)$

So, a = ${{{1 + \sqrt {17} } \over 2}}$

${\sin ^{ - 1}}{{12} \over {13}} - {\sin ^{ - 1}}{3 \over 5} = {\sin ^{ - 1}}\left( {{{12} \over {13}}.{4 \over 5}.{3 \over 5}.{5 \over {13}}} \right)$

$ \Rightarrow {\sin ^{ - 1}}{{33} \over {65}} = {\pi \over 2} - {\cos ^{ - 1}}{{33} \over {65}}$

$ \Rightarrow {\pi \over 2} - {\sin ^{ - 1}}{{56} \over {65}}$

${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $

$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha $

$ \Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha $

$\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} $

squaring both sides

${x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha $

$ \therefore $ 4x² – 4xy cos $\alpha $ + y² = 4 ${\sin ^2}\alpha $

Here $\cos \alpha = {3 \over 5}$

$ \therefore $ $\tan \alpha = {4 \over 3}$

and $\tan \beta = {1 \over 3}$

We know,

$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$

= ${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$ = ${9 \over {13}}$

$ \therefore $ $\left( {\alpha - \beta } \right)$ = ${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$ = ${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$

tan^$-$1(2x) + tan^$-$1(3x) = $\pi $/4

$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$ = 1

$ \Rightarrow $  6x² + 5x $-$ 1 = 0

x = $-$1 or x = ${1 \over 6}$

x = ${1 \over 6}$

$ \because $  x > 0

cot^$-$1 x > 5,    cot^$-$1 x < 2

$ \Rightarrow $  x < cot5,   x > cot2

$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)$

$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$

$\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} $

$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}$

(Where tanA $=$ 20, tanB $=$ 1)   ${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$

x = sin^$-$1 sin 10 = 3$\pi $ $-$ 10

y = cos^$-$1cos 10 = 4$\pi $ $-$ 10

y $-$ x = (4$\pi $ $-$ 10) $-$ (3$\pi $ $-$ 10) = $\pi $

Given,

${\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) + {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right) = {\pi \over 2}$

$ \Rightarrow {\cos ^{ - 1}}\left( {{2 \over {3x}}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {{3 \over {4x}}} \right)$

$ \Rightarrow \cos \left( {{{\cos }^{ - 1}}\left( {{2 \over {3x}}} \right)} \right) = \cos \left[ {{\pi \over 2} - {{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right]$

$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\cos }^{ - 1}}\left( {{3 \over {4x}}} \right)} \right\}$

$ \Rightarrow {2 \over {3x}} = \sin \left\{ {{{\sin }^{ - 1}}{{\sqrt {16{x^2} - 9} } \over {4x}}} \right\}$

$ \Rightarrow {2 \over {3x}} = {{16{x^2} - 9} \over {4x}}$

$ \Rightarrow 64 = 9\left( {16{x^2} - 9} \right)$

$ \Rightarrow 16{x^2} - 9 = {{64} \over 9}$

$ \Rightarrow 16{x^2} = {{64} \over 9} + 9$

$ \Rightarrow 16{x^2} = - {{145} \over 9}$

$ \Rightarrow x = \pm {{\sqrt {145} } \over {4 \times 3}}$

$ \Rightarrow x = \pm {{\sqrt {145} } \over {12}}$

as given that $x > {3 \over 4}$

$ \therefore $  x $ \ne $ $ - {{\sqrt {145} } \over {12}}$

$ \therefore $  x $ = {{\sqrt {145} } \over {12}}$

Let, ${\cot ^{ - 1}}(1 + x) = \alpha $

$ \Rightarrow 1 + x = \cot \alpha $

Now, let ${\tan ^{ - 1}}(x) = \beta $

$ \Rightarrow x = \tan \beta $

Given, $\sin ({\cot ^{ - 1}}(x)) = \cos ({\tan ^{ - 1}}(x))$

$ \Rightarrow \sin \alpha = \cos \beta $

From $\Delta$ABC, $\sin \alpha = {1 \over {\sqrt {1 + {x^2} + 2x + 1} }}$

$ = {1 \over {\sqrt {{x^2} + 2x + 2} }}$

From $\Delta$MNO, $\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}$

$\therefore$ ${1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}$

$ \Rightarrow {x^2} + 2x + 2 = {x^2} + 1$

$ \Rightarrow 2x + 2 = 1$

$ \Rightarrow 2x = - 1$

$ \Rightarrow x = -{1 \over 2}$

Given,

tan^-1 $\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$

Let    x² = cos $\theta $

= tan^-1 $\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$

= tan^-1 $\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$

= tan^-1 $\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$

= tan^-1 $\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$

= tan^-1 $\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$

= tan^-1 $\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$

= tan^-1 $\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$

= ${\pi \over 4} + {\theta \over 2}$

= ${\pi \over 4} + {1 \over 2}$ cos^-1 x²

Given,

${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right)$

$ \Rightarrow {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{x + {{2x} \over {1 - {x^2}}}} \over {1 - x\left( {{{2x} \over {1 - {x^2}}}} \right)}}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\tan ^{ - 1}}\left( {{{x - {x^3} + 2x} \over {1 - {x^2} - 2{x^2}}}} \right)$

$\therefore$ $\,\,\,$ ${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$

$ \Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$

Given that, $x,y,z\,\,$ are in $AP$

So, $\,\,\,$ $2y = x + y$

Also given that,

${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$ and $\,\,\,{\tan ^{ - 1}}z\,\,$ are in $AP$

So, $2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$

$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)$

$ \Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$

$ \Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$

[ as $\,\,\,\,$ $2y = x + z$]

$ \Rightarrow 1 - {y^2} = 1 - xz$

$ \Rightarrow $ ${y^2} = xz$

As we get ${y^2} = xz,$ so, $x,y,z$ are in $GP.$

According to the question $x,y,z$ are $AP.$

$x,y,z$ both can be $AP$ as well as $GP$

when $x=y=z.$

Given,

$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$

$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$

$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$

$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{17} \over 6}} \right)} \right)$

$ = \cot \left( {{{\cot }^{ - 1}}{6 \over {17}}} \right)$

$ = {6 \over {17}}$

Given sin^-1$\left( {{x \over 5}} \right)$ + cosec^-1$\left( {{5 \over 4}} \right)$ = ${\pi \over 2}$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ + sin^-1$\left( {{4 \over 5}} \right)$ = ${\pi \over 2}$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ = ${\pi \over 2}$ - sin^-1$\left( {{4 \over 5}} \right)$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ = cos^-1$\left( {{4 \over 5}} \right)$

$ \Rightarrow $ ${x \over 5}$ = sin(cos^-1$ {{4 \over 5}}$)

$ \Rightarrow $ ${x \over 5}$ = sin(sin^-1$ {{3 \over 5}}$)

$ \Rightarrow $ ${x \over 5}$ = ${3 \over 5}$

$ \Rightarrow $ x = 3

As we know,

${\cos ^{ - 1}}A - {\cos ^{ - 1}}B$

$ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)$

Given, ${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $

$ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha $

$ \Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x$

$ \Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)$

$ \Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}$

$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x$

$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x$

$ \Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x$

Given that,

${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$

We know,

$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$

$\therefore$ $\,\,\,$ $ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$

$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$

$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$

$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$

$\therefore$ $\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$

Given that,

${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$

We know,

${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$

$\therefore$ $\,\,\,$ ${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$

Adding $(1)$ and $(2),$ we get,

$2{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = x + {\pi \over 2}$

$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$

$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$

$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$

$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$

Squaring both sides we get,

$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$

$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$

$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$

Applying compounds and dividendo rule,

$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$

$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$

Other Method :

$ \begin{aligned} & \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned} $