sin x = -1
\( \sin x=-\sin \frac{\pi}{2}=\sin \left(-\frac{\pi}{2}\right) \)
\(x=(4 n-1) \frac{\pi}{2}, n=0, \pm 1\)

sin^-1[sin10]
We know that sin^-1(sin \(\theta\)) = \(\theta\) is \(\theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)Considering the approximation \(\frac{\pi}{2}=\frac{11}{7}\)
we conclude that 10\(\notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \), but (10-3\(\pi\)) \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \).
Now, sin10 = sin(3\(\pi\)+(10−3\(\pi\))) = sin(\(\pi\)+(10−3\(\pi\)) = −sin(10−3\(\pi\)) = sin(3\(\pi\)-10)
Hence, sin^-1[sin10] = sin^-1[sin(3\(\pi\)-10)] = 3\(\pi\)-10, since(3\(\pi\)-10)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \).

\({ sec }^{ -1 }\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) \)
Note that \(\frac{5\pi}{3}\) is not in [0, \(\pi\)]\{\(\frac{\pi}{2}\)}, the principal range of sec^-1 x.
we write \(\frac { 5\pi }{ 3 } =2\pi -\frac { \pi }{ 3 } \).
Now, sec\(\left( \frac { 5\pi }{ 3 } \right) =sec\left( 2\pi -\frac { \pi }{ 3 } \right) =sec\left( \frac { \pi }{ 3 } \right) and\frac { \pi }{ 3 } \in [0,\pi ]\)\{\(\frac{\pi}{2}\)}
Hence, sec^-1\(\left( sec\left( \frac { 5\pi }{ 3 } \right) \right) ={ sec }^{ -1 }\left( sec\left( \frac { \pi }{ 3 } \right) \right) =\frac { \pi }{ 3 } \)

tan^-1\((tan(\frac{3\pi}{4})\)
Observe that \(\frac{3\pi}{4}\) is not in the interval \(\left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \), the principal range of tan^-1 x.
So, we write \(\frac{3\pi}{4}=\pi-\frac{\pi}{4}\)
Now, \(tan\left( \frac { 3\pi }{ 4 } \right) =tan\left( \pi -\frac { \pi }{ 4 } \right) =-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) and-\frac { \pi }{ 4 } \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
Hence, \({ tan }^{ -1 }\left( tan\left( \frac { 3\pi }{ 4 } \right) \right) ={ tan }^{ -1 }\left( tan\left( -\frac { \pi }{ 4 } \right) \right) =-\frac { \pi }{ 4 } ,since-\frac { \pi }{ 4 } \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

\({ sin }^{ -1 }(\frac { 3 }{ 5 } )-{ cos }^{ -1 (}\frac { 12 }{ 13 } )={ sin }^{ -1 }(\frac { 16 }{ 65 }) \)
Let \({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\Rightarrow sinx=\frac { 3 }{ 5 } \)

and \(cosx=\frac { adj }{ hyp } =\frac { 4 }{ 5 } \)
and \({ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) =y\Rightarrow cosy=\frac { 12 }{ 13 } \)
\(siny=\frac { opp }{ hyp } =\frac { 5 }{ 12 } \)

We know that
sin (x - y) = sin x cos y -cos x siny
= \(\frac { 3 }{ 5 } \times \frac { 12 }{ 13 } -\frac { 4 }{ 5 } \times \frac { 5 }{ 12 } =\frac { 36 }{ 65 } -\frac { 20 }{ 6g } =\frac { 16 }{ 65 } \)
\(\therefore x-y={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
\(\Rightarrow { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) -{ cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) ={ sin }^{ -1 }\left( \frac { 16 }{ 65 } \right) \)
Hence proved

\(tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) \)
Let \(sin^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\)

\(\Rightarrow \frac { 3 }{ 5 } =sinx\)
\(\therefore tanx=\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\({ cot }^{ -1 }\left( \frac { 3 }{ 2 } \right) =y\)
\(\Rightarrow \frac { 3 }{ 2 } =coty\Rightarrow tany=\frac { 2 }{ 3 } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =tan(x+y)\)
\(\frac { tanx+tany }{ 1-tanxtany } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\frac { \frac { 3 }{ 2 } +\frac { 2 }{ 3 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 2 }{ 3 } \right) } =\frac { \frac { 9+6 }{ 12 } }{ 1-\frac { 6 }{ 12 } } \)
\(\therefore tan\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ cot }^{ -1 }\frac { 3 }{ 2 } \right) =\frac { 17 }{ 6 } \)

 \(cot\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ sin }^{ -1 }\frac { 4 }{ 5 } \right) \)
\({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) =x\Rightarrow sinx=\frac { 3 }{ 5 } \)
\(\therefore tanx=\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\({ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) =y\Rightarrow siny=\frac { 4 }{ 5 } \)
\(\therefore tany=\frac { opp }{ adj } =\frac { 4 }{ 3 } \)

\(\therefore cot\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ sgn }^{ -1 }\frac { 4 }{ 5 } \right) =cot\left( x+y \right) \)
= \(\frac { 1 }{ tan(x+y) } =\frac { 1 }{ \frac { tanx+tany }{ 1-tanxtany } } =\frac { 1-tabxtany }{ tanxetany } \)


\(\therefore cot\left( { sin }^{ -1 }\frac { 3 }{ 5 } +{ sin }^{ -1 }\frac { 4 }{ 5 } \right) =0\)

tan\(\left( { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) \right) \)

We know that \(sin^{ -1 }\left( x \right) ={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \) if-1 \(\therefore { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) ={ tan }^{ -1 }\left( \frac { x\frac { 1 }{ 2 } }{ 1-\left( x+\frac { 1 }{ 2 } \right) ^{ 2 } } \right) \)
= \({ tam }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { 1-{ x }^{ 2 }-\frac { 1 }{ 4 } -x } } \right) \)
= \({ tan }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { { x }^{ 2 }-x+\frac { 3 }{ 4 } } } \right) \)
\(\therefore tan\left( { sin }^{ -r }\left( x+\frac { 1 }{ 2 } \right) \right) =tan\left( { tan }^{ -1 }\left( \frac { x+\frac { 1 }{ 2 } }{ \sqrt { -{ x }^{ 2 }-x+\frac { 3 }{ 4 } } } \right) \right) \)
= \(\frac { x+\frac { 1 }{ 2 } }{ \sqrt { -{ x }^{ 2 }-x+\frac { 3 }{ 4 } } } =\frac { 2 }{ \frac { \sqrt { { -4x }^{ 2 }-4x+3 } }{ 2 } } \)
\(tan\left( { sin }^{ -1 }\left( x+\frac { 1 }{ 2 } \right) \right) =\frac { 2x+1 }{ \sqrt { 3+4x-{ 4x }^{ 2 } } } \)

cos (tan^-1(3x-1))

We know that \({ tan }^{ -1 }\left( x \right) ={ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) ifx\ge 0\)
\({ tan }^{ -1 }\left( 3x-1 \right) ={ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 1+(3x-1)^{ 2 } } } \right) \)
\(\Rightarrow { cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 1+9x+1-6x } } \right) ={ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { { 9x }^{ 2 }-6x+2 } } \right) \)
\(\therefore cos\left( { tan }^{ -1 }\left( 3x-1 \right) \right) =cos\left( { cos }^{ -1 }\left( \frac { 1 }{ \sqrt { { 9 }x^{ 2 }-6x+2 } } \right) \right) \)
= \(\frac { 1 }{ \sqrt { { 9x }^{ 2 }-6x+2 } } \)

\(cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \right) \)
\(\Rightarrow \frac { 4 }{ 5 } =sinx\)

\(\therefore cosx=\frac { adj }{ hyp } =\frac { 3 }{ 5 } \)
Let \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) =y\)
\(\Rightarrow tany=\frac { 3 }{ 4 } \)

\(\Rightarrow siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \ cosy=\frac { adj }{ hup } =\frac { 4 }{ 5 } \)
\(\therefore cos\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) -{ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
= cos (x - y) = cos x cos y + sin x sin y
= \(\frac { 3 }{ 5 } .\frac { 4 }{ 5 } +\frac { 4 }{ 5 } .\frac { 3 }{ 5 } \)
= \(\frac { 12 }{ 25 } +\frac { 12 }{ 25 } =\frac { 24 }{ 25 } \)
= \(\frac { 24 }{ 25 } \)

\(sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)

Let \({ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\Rightarrow \frac { 1 }{ 2 } =tanx\)
\(cosx=\frac { adj }{ hyp } =\frac { 2 }{ \sqrt { 5 } } \)
\(\therefore sinx= \frac { 1 }{ \sqrt { 5 } } \)

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =y\Rightarrow \frac { 4 }{ 5 } =cosy\)
\(siny=\frac { opp }{ hyp } =\frac { 3 }{ 5 } \)
\(\therefore sin\left( { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) -{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
= sin (x - y) = sin x cos y - cos x sin y
= \(\frac { 1 }{ \sqrt { 5 } } .\frac { 4 }{ 5 } -\frac { 2 }{ \sqrt { 5 } } .\frac { 3 }{ 5 } \)
= \(\frac { 4 }{ 5\sqrt { 5 } } =\frac { 6 }{ 5\sqrt { 5 } } =\frac { -2 }{ 5\sqrt { 5 } }\)

\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) .\)
\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) ={ cos }^{ -1 }\left( cos\left( 2\pi -\frac { 2\pi }{ 3 } \right) \right) \)
[\(\therefore\) the period of cosine in 2\(\pi \)]
= \({ cos }^{ -1 }\left( cos\frac { 2\pi }{ 3 } \right) \) \(\left[ \because cos\left( 2\pi -\theta \right) =cos\theta \right] \)
= \(\frac { 2\pi }{ 3 } \) \(\left[ \because \frac { 2\pi }{ 3 } \epsilon \left[ 0,\pi \right] \right] \)
\({ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) ={ cop }^{ -1 }\left( cos\left( 2\pi -\frac { 3\pi }{ 4 } \right) \right) \) = [ \(\therefore\) Period of cosine is 2\(\pi\)]
 \({ cos }^{ -1 }\left( cos\left( 3\frac { \pi }{ 4 } \right) \right) \)  \(\left[ \because cos\left( 2\pi -\theta \right) =cos\theta \right] \)
= \(3\frac { \pi }{ 4 } \) \(\left[ \because 3\frac { \pi }{ 4 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\therefore \) \({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) \)
= \(\frac { 2\pi }{ 3 } +3\frac { \pi }{ 4 } \)
= \(\frac { 8\pi +9\pi }{ 12 } =\frac { 7\pi }{ 1b } \)
\({ cos }^{ -1 }\left( cos\left( \frac { 4\pi }{ 3 } \right) \right) +{ cos }^{ -1 }\left( cos\left( \frac { 5\pi }{ 4 } \right) \right) \) = \(\frac { 7\pi }{ 12 } \)

g(x) = 2sin⁻¹(2x−1)−\(\frac{\pi}{4}\)
From the definition of sin^-1x,
\(-1\le 2x-1\le 1\)
\(\Rightarrow 1+1\le 2x\le 1+1\)
\(\Rightarrow 0\le 2x\le 2\)
\(\Rightarrow 0\le x\le 1\)
\(\therefore \) Domain = [0, 1]

\({ tan }^{ -1 }(\frac { 2 }{ 11 }) +{ tan }^{ -1 }(\frac { 7 }{ 24 }) ={ tan }^{ -1 }(\frac { 1 }{ 2 } )\)
\(LHS={ tan }^{ -1 }\left( \frac { 2 }{ 11 } \right) +{ tan }^{ -1 }\left( \frac { 7 }{ 24 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 11 } +\frac { 7 }{ 24 } }{ 1-\left( \frac { 2 }{ 11 } \right) \left( \frac { 7 }{ 24 } \right) } \right) \) \(\left[ \because { tan }^{ -1 }(x)+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { \frac { 48+77 }{ 11\times 24 } }{ \frac { 264-14 }{ 264 } } \right) ={ tan }^{ -1 }\left( \frac { \frac { 125 }{ 264 } }{ \frac { 250 }{ 264 } } \right) \)

= RHS
Hence proved.

\({ sin }^{ -1 }\left( cos\left( { sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \right) \right) \)
Let \({ sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) =x\)
\(\Rightarrow \frac { \sqrt { 3 } }{ 2 } =sinx\)
\(\Rightarrow sinx=sin\frac { \pi }{ 3 } \)  \(\left[ \because \frac { \pi }{ 3 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ -1 }\left( cbs\left( { sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \right) \right) ={ sin }^{ -1 }\left( cos\left( \frac { \pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
Let \({ sin }^{ -1 }\frac { \pi }{ 6 } =siny\)
\(\Rightarrow y=\frac { \pi }{ 6 } \quad \left[ \because \frac { \pi }{ 6 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\therefore { sin }^{ -1 }\left( cos\left( { sin }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) \right) \right) =\frac { \pi }{ 6 } \)

sin(cos⁻¹(1-x))
we know that \({ cos }^{ -1 }x={ sin }^{ -1 }\left( \sqrt { 1-{ x }^{ 2 } } \right) \text {if}\ 0\le x\le 1\)
\(\therefore { cos }^{ -1 }\left( 1-x \right) ={ sin }^{ 1 }\sqrt { 1-\left( 1-x \right) ^{ 2 } } \left[ \because 0\le x\le 1 \right] \)
= \({ sin }^{ -1 }\left( \sqrt { 1-\left( 1+{ x }^{ 2 }-2x \right) } \right) \)
= \({ sin }^{ -1 }\left( \sqrt { 1-1-{ x }^{ 2 }+2x } \right) ={ sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \)
\(\therefore sin\left( { cos }^{ -1 }\left( 1-x \right) \right) =sin\left( { sin }^{ -1 }\left( \sqrt { 2x-{ x }^{ 2 } } \right) \right) \)
= \(\sqrt { 2x-{ x }^{ 2 } } \)

\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) \).
The range of principal values of cos^-1x is [0, \(\pi\)].
Since \(\frac{13\pi}{3}\not \in[0,\pi]\), we write \(\frac{13\pi}{3} as \frac{13\pi}{3}=4\pi+\frac{\pi}{3}, whre \frac{\pi}{3}\in[0,\pi]\)
Now, cos\((\frac{13\pi}{3}=cos (4\pi+\frac{\pi}{3})=cos\frac{\pi}{3}\)
Thus,\({ cos }^{ -1 }\left( cos\left( \frac { 13\pi }{ 3 } \right) \right) ={ cos }^{ -1 }\left( cos\left( \frac { \pi }{ 3 } \right) \right) ,\ since\frac { \pi }{ 3 } \in [0,\pi ]\).

Let cot−1\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) =\alpha \). Then, cot \(\alpha =\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \) and α is acute.
We construct a right triangle with the given data.
From the triangle, sec\(\alpha=\frac{x}{1}=x\). Thus, \(\alpha\) = sec^-1x
Hence, cot⁻¹\(\left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ={ sec }^{ -1 }x,|x|>1\)

 \(tan\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right) \)
Let \({ cos }^{ -1 }\left( \frac { -1 }{ 2 } \right) =x\)
\(\Rightarrow \frac { 1 }{ 2 } =cosx\)
\(\Rightarrow cosc=cos\frac { \pi }{ 3 } \)
\(\Rightarrow x=\frac { \pi }{ 3 } \)
Let \({ sin }^{ -1 }\left( \frac { -1 }{ 2 } \right) =y\)
\(\Rightarrow \left( \frac { -1 }{ 2 } \right) =siny\)
\(\Rightarrow siny=\frac { -1 }{ 2 } =-sin\frac { \pi }{ 6 } =\left( \frac { -\pi }{ 6 } \right) \)
\(\Rightarrow y=\frac { -\pi }{ 6 } \)
\(\therefore { tan }^{ -1 }\left( cos^{ -1 }\left( \frac { 1 }{ 2 } \right) -{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \({ tan }^{ -1 }\left( \frac { \pi }{ 3 } -\left( \frac { -\pi }{ 6 } \right) \right) ={ tan }^{ -1 }\left( \frac { \pi }{ 3 } +\frac { \pi }{ 6 } \right) \)
= \(tan\left( \frac { 2\pi +\pi }{ 0 } \right) =tan\left( \frac { 3\pi }{ 6 } \right) =tan\left( \frac { \pi }{ 2 } \right) \)
= \(\infty \)

By definition, the domain of yx = cos^-1 x is -1. This leads to \(-1\le\frac{2+sinx}{3}\le1\) which is same as -3\(\le\)2+sinx\(\le\)3
so, -5\(\le sin\ x\le1 \) reduces to -1\(\le sin\ x\le1 \), which gives
-sin^-1(1)\(\le x\le sin^-1(1) or -\frac{\pi}{2}\le x\le \frac{\pi}{2}\)
Thus, the domain of cos-1\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

\(={ sin }^{ -1 }\left( sin\frac { 5\pi }{ 9 } cos\frac { \pi }{ 9 } +cos\frac { 5\pi }{ 9 } sin\frac { \pi }{ 9 } \right) \)
= \({ sin }^{ - }\left( sin\left( \frac { 5\pi }{ 9 } +\frac { \pi }{ 9 } \right) \right) \)
(\(\because \) sin A cos B + cos A sin B = sin (A + B))
= \({ sin }^{ -1 }\left( sin\left( \frac { 6\pi }{ 9 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \frac { 2\pi }{ 3 } \right) \right) \)
= \({ sin }^{ -1 }\left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) \) \(\left[ \because \frac { 2\pi }{ 3 } \notin \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
= \({ sin }^{ -1 }\left( sin\frac { \pi }{ 3 } \right) \) \(\left( \because sin\left( \pi -\theta \right) =sin\theta \right) \)
= \(\frac { \pi }{ 3 } \) \(\left[ \because \frac { \pi }{ 3 } \quad \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \) 

Given \(f(x)=sin^{ -1 }\left( \frac { x^{ 2 }+1 }{ 2x } \right) \le 1\)
We know that the domain of sin^-1(x) is [-1, 1]
\(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \le 1\)
Consider \(\Rightarrow -1\le \frac { { x }^{ 2 }+1 }{ 2x } \)
\(\Rightarrow 0\le \frac { { x }^{ 2 }+1 }{ 2x } +1\)
\(\Rightarrow \frac { { x }^{ 2 }+1+2x }{ 2x } \ge 0\)
\(\Rightarrow \frac { \left( x+1 \right) ^{ 2 } }{ 2x } \ge 0\)
\(\Rightarrow \) x = -1 and x < 0
Consider \(\cfrac { { x }^{ 2 }+1 }{ 2x } \le 1\)
\(\Rightarrow \frac { { x }^{ 2 }+1 }{ 2x } -1\le 0\)
\(\Rightarrow \frac { { x }^{ 2 }-2x+1 }{ 2x } \le 0\)
\(\Rightarrow \frac { \left( x-1 \right) ^{ 2 } }{ 2x } \le 0\)
From (1) and (2) Domain {-1, 1}

Given sin x = 0
\(\Rightarrow\) sin x = sin 0
\(\Rightarrow\) \(​​x=n\pi ,n\varepsilon z\)
Since \(-10\pi \le x\le 10\pi \) n can take the values only from -10 to +10.
\(\therefore\) \(x=n\pi ,\) When \(n=0,\pm ,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 7,\pm 8,\pm 9,\pm 10\)

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

\({ tan }^{ -1 }\left( tan\frac { 3\pi }{ 5 } \right) \)
Let us find \(\theta \in(-\frac{\pi}{2},\frac{\pi}{2})\) such that tan \(\theta\) = tan\(\frac{3\pi}{5}\)
Since the tangent function has period \(\pi, tan \frac{3\pi}{5}=tan(\frac{3\pi}{5}-\pi)=tan(-\frac{2\pi}{5})\)
Therefore,  \({ tan }^{ -1 }\left( tan\frac { 3\pi }{ 5 } \right) ={ tan }^{ -1 }\left( tan\left( -\frac { 2\pi }{ 5 } \right) \right) =-\frac { 2\pi }{ 5 } ,since\frac { -2\pi }{ 5 } \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) .\)

\({ tan }^{ -1 }\left( -\sqrt { 3 } \right) ={ tan }^{ -1 }\left( tan\left( -\frac { \pi }{ 3 } \right) \right) =-\frac { \pi }{ 3 } \)
since \(-\frac { \pi }{ 3 } \in \left( \frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

Since tan(tan^-1 x) = x, x ∈ R,
We have tan(tan^-1(2019)) = 2019

\(cos\left( { cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
\({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow \frac { 4 }{ 5 } =cos\theta \)
Also \({ sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ sin }^{ -1 }\left( cos\theta \right) \) [using (1)]
= \({ sim }^{ -1 }\left( sin\left( \frac { \pi }{ 2 } -\theta \right) \right) \) \(\left[ \because cos\theta =sin\left( \frac { \pi }{ 2 } -\theta \right) \right] \)
= \(\frac { \pi }{ 2 } -\theta \)
\(\therefore cos\left( cos^{ i1 }\left( \frac { 4 }{ 5 } \right) +sin^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) \)
= \(cos\left( \theta +\frac { \pi }{ 2 } -\theta \right) \)[using (1) & (2)]
= \(cos\frac { \pi }{ 2 } \)
= 0

\({ cot }^{ -1 }x-{ xot }^{ -1 }\left( x+2 \right) =\frac { \pi }{ 12 } ,x>0\)
\({ tan }^{ -1 }\left( \frac { 1 }{ x } \right) -{ tan }^{ -1 }\left( \frac { 1 }{ x+2 } \right) =\frac { \pi }{ 2 } \)
\(\left[ \because { cot }^{ 1 }\left( x \right) ={ tan }^{ -1 }\left( \frac { 1 }{ x } \right) ifx>0 \right] \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { \frac { 1 }{ x } +\frac { 1 }{ x+2 } }{ 1+\frac { 1 }{ x } .\frac { 1 }{ x+2 } } \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow \left( \frac { \frac { x+2-x }{ x(x+2) } }{ \frac { x(x+2)+1 }{ x(x+2) } } \right) ={ tan15 }^{ 0 }\)
\(\left[ \because { tan15 }^{ 0 }=tan\left( { 45 }^{ 0 }-30^{ 0 } \right) \right] \)
\(\frac { { tan45 }^{ 0 }-{ tan30 }^{ 0 } }{ 1+tan{ 45 }^{ 0 }tan{ 30 }^{ 0 } } \)
\(\frac { 1-\frac { 1 }{ \sqrt { 3 } } }{ 1+\frac { 1 }{ \sqrt { 3 } } } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
\(\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 3-1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)

\(\Rightarrow \frac { 2 }{ { x }^{ 2 }+2x+1 } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow \frac { 7 }{ (x+1)^{ 2 } } =\frac { \left( \sqrt { 3 } -1 \right) ^{ 2 } }{ 2 } \)
\(\Rightarrow 4=(x+1)^{ 2 }\left( \sqrt { 3 } -1 \right) ^{ 2 }\)
Taking square root both sides
\(2=(x+1)(\sqrt { 3 } -1)\)
\(\Rightarrow x+1=\frac { 2 }{ \sqrt { 3- } 1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 2\left( \sqrt { 3 } +1 \right) }{ 3-1 } =\sqrt { 3 } +1\)
\(x+1=\sqrt { 3 }+ 1\)
\(x=\sqrt { 3 } \)

\(2{ tan }^{ -1 }(cosx)={ tan }^{ -1 }(2cosec\ x)\)
Given \({ 2tan }^{ -1 }(cosx)\quad \left[ \because { tan }^{ -1 }+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
Let us find 2 tan(cos x)
= \({ 2tan }^{ -1 }(cosx)\ \left[ \because { tan }^{ -1 }+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \frac { cosx+cosx }{ 1-{ cos }^{ 2 }x } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) \)
\(2{ tan }^{ -1 }(cosx)={ tan }^{ -1 }\left( 2cosecx \right) =0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) -{ tan }^{ -1 }(2cosecx)=0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ { sin }^{ 2 }x } \right) -{ tan }^{ -1 }\left( \frac { 2 }{ sinx } \right) =0\)
\(\Rightarrow { tan }^{ -1 }\left( \frac { \frac { 2cosx }{ { sin }^{ 2 }x } -\frac { 2B }{ sinx } }{ 1+\frac { 2cosx }{ { sin }^{ 2 }x } .\frac { 2 }{ sinx } } \right) =0\)
\(\Rightarrow 2sin\ x\ cosx-2{ sin }^{ 2 }x=0\)
\(\Rightarrow 2sin\ x\left( cosx-sinx \right) =0\)
\(\Rightarrow sinx=0\ or\ cosx=sinx\)
\(\Rightarrow sinx=0\ or\ tanx-1\)
\(\Rightarrow sinx=0\ or\ tanx=tan\cfrac { \pi }{ 4 } \)
The solutions for sin x = 0 is \(x=n\pi ,\theta \varepsilon zd\)  and the solution for
\(tan\ x=tan\frac { \pi }{ 4 } \ x=n\pi +\frac { \pi }{ 4 } ,n\varepsilon Z\)
\(\left[ \therefore tan\ x=tan\ \alpha \Rightarrow tan\ x=2n\pi +\alpha ,n\varepsilon Z \right] \)
Hence, the solutions are \(x=n\pi \) or 
\(x=n\pi +\frac { \pi }{ 4 } ,n\varepsilon z.\)

\(2{ tan }^{ -1 }x={ cos }^{ -1 }\frac { 1-{ a }^{ 2 } }{ 1+{ a }^{ 2 } } -{ cos }^{ -1 }\frac { 1-{ b }^{ 2 } }{ 1+{ b }^{ 2 } } ,a>0,b>0\)
Let \(a=tan\ \theta \ b=tan\phi \)
\(\therefore { cos }^{ -1 }\left( \frac { 1-{ a }^{ 2 } }{ 1+a^{ 2 } } \right) ={ cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right) \)
= \({ cos }^{ -1 }\left( cos2\theta \right) =2\theta \)    ...(1)
\(\left[ \because cos2\theta =\frac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \right] \)
Also \({ cos }^{ -1 }\left( \frac { 1-{ b }^{ 2 } }{ 1+{ b }^{ 2 } } \right) ={ cos }^{ -1 }\left( \frac { 1-{ tan }^{ 2 }\phi }{ 1+tan^{ 2 } } \right) \)
= \(cos^{ -1 }\left( cos2\Phi \right) \)
\(\therefore 2{ tan }^{ -1 }x=2\theta -2\phi =2(\theta -\phi )\)
[using (1) and (2)]
\(\Rightarrow tan^{ -1 }x=\theta -\phi \)
\(\Rightarrow x=tan\left( \theta -\phi \right) =\frac { tan\theta -tan\phi }{ 1+tan\theta tan\phi } \)
\(\left[ \because tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB } \right] \)
\(\Rightarrow x=\frac { a-b }{ 1+ab } \), a > 0

\({ sin }^{ -1 }\frac { 5 }{ x } +{ sin }^{ -1 }\frac { 12 }{ x } =\frac { \pi }{ 2 } \)
We know that
\({ sin }^{ -1 }x+{ sin }^{ -1 }y=sin^{ -1 }\left( x\sqrt { 1-{ y }^{ 2 } } +y\sqrt { 1-{ x }^{ 2 } } \right) \)
where either x²+y²\(\pm \)1 or xy<0
\({ sin }^{ -1 }\left[ \left( \frac { 5 }{ x } \sqrt { 1-\frac { { 12 }^{ 2 } }{ { x }^{ 2 } } } \right) +\frac { 12 }{ x } \sqrt { 1-\frac { { 5 }^{ 2 } }{ { x }^{ 2 } } } \right] =\frac { \pi }{ 2 } \)
This holds true only either
Let \(x=\frac { 5 }{ x } \quad y=\frac { 12 }{ x } \)
\(\therefore { x }^{ 2 }+{ y }^{ 2 }\le 1\Rightarrow \frac { { 5 }^{ 2 } }{ { x }^{ 2 } } +\frac { { 12 }^{ 2 } }{ { x }^{ 2 } } \le 1\Rightarrow \frac { 25 }{ { x }^{ 2 } } +\frac { 144 }{ { x }^{ 2 } } \le 1\)
\(\Rightarrow \frac { 169 }{ { x }^{ 2 } } \le 1\Rightarrow 169\le { x }^{ 2 }\Rightarrow { x }^{ 2 }\ge 169\)
\(\Rightarrow x\ge \sqrt { 169 } \Rightarrow x\ge 13\quad or\quad \left( \frac { 5 }{ x } \right) \left( \frac { 12 }{ x } \right) <0\)
which is not possible,
\(\therefore x\ge 13\)
Given \({ sin }^{ -1 }\left( \frac { 5 }{ x } \right) +{ sin }^{ -1 }\left( \frac { 12 }{ x } \right) =\frac { \pi }{ 2 } \)
\(\Rightarrow { sin }^{ -1 }\left( \frac { 5 }{ x } \right) =\left( \frac { \pi }{ 2 } \right) -{ sin }^{ -1 }\left( \frac { 12 }{ x } \right) \)
\(\Rightarrow { sin }^{ -1 }\left( \frac { 5 }{ x } \right) ={ cos }^{ -1 }\left( \frac { 12 }{ x } \right) \)
\(\left[ \because { sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } \right] \)
\(\left[ \therefore { sin }^{ -1 }\left( \frac { 5 }{ 13 } \right) =A\Rightarrow sinA=\frac { 5 }{ 13 } \Rightarrow cosA=\frac { 12 }{ 13 } Let\quad { cos }^{ -1 }\left( \frac { 12 }{ 13 } \right) =B\Rightarrow cosB=\frac { 12 }{ 13 } \right] \)
This is true only when x = 13
\(\therefore\) x = \(\pm \)13 is the solution

We know that \(sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) =cos^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
Thus, \(cos\left( sin^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right) =\frac { 1 }{ \sqrt { 1+x^{ 2 } } } \) ...(1)
Let \(\cot ^{-1}\left(\frac{3}{4}\right)=\theta\). Then \(\cot \theta=\frac{3}{4}\) and so \(\theta\) is cute. 
From the diagram, we get,
Hence \(\sin \left\{\cot ^{-1}\left(\frac{3}{4}\right)\right\}=\sin \theta=\frac{4}{5}\) ................ (2)
Using (1) and (2) in the given equation, we \(\frac { 1 }{ \sqrt { 1+x^{ 2 } } } =\frac { 4 }{ 5 } \) \(\sqrt{1+x^2}=\frac{5}{4}\)
Thus, x = \(\pm\frac{3}{4}\)

Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

Let sec^-1 \(\frac{5}{4}=\theta\).
Then, sec \(\theta=\frac{5}{4}\) and hence, cos \(\theta=\frac{4}{5}\)
Also, sin \(\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-{ \left( \frac { 4 }{ 5 } \right) }^{ 2 } } =\frac { 3 }{ 5 } \), which gives \(\theta=sin^{-1}(\frac{3}{5})\)
Thus, sec^-1\(\left( \frac { 5 }{ 4 } \right) ={ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) and\ { sin }^{ -1 }\frac { 3 }{ 5 } +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) =2{ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
We know that sin^-1\(\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) =2{ sin }^{ -1 }x,\quad if|x|\le \frac { 1 }{ \sqrt { 2 } } \)
Since \(\frac { 3 }{ 5 } <\frac { 1 }{ \sqrt { 2 } } \), we have 2\({ sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) ={ sin }^{ -1 }\left( 2\times \frac { 3 }{ 5 } \sqrt { { 1-\left( \frac { 3 }{ 5 } \right) }^{ 2 } } \right) ={ sin }^{ -1 }\left( \frac { 24 }{ 25 } \right) \)
Hence, \(sin\left[ { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) +{ sec }^{ -1 }\left( \frac { 5 }{ 4 } \right) \right] =sin\left( { sin }^{ -1 }\left( \frac { 24 }{ 25 } \right) \right) =\frac { 24 }{ 25 } ,\ since\frac { 24 }{ 25 } \in [-1,1]\)

If x = 0 , then both sides are equal to 0  .... (1)
Assume that 0< x <1.
Let \(\theta\) = sin⁻¹ x. Then 0 < \(\theta<\frac{\pi}{2}\). Now, sin \(\theta=\frac{x}{1}\) gives tan \(\theta=\frac{x}{\sqrt{1-x^2}}\)
Hence, tan (sin^-1 x) = \(\frac{x}{\sqrt{1-x^2}}\) ....... (2)
Assume that −1 < x < 0. Then, \(\theta\) = sin⁻¹ x gives −\(\frac{\pi}{2}<\theta<0.\) Now, sin \(\theta=\frac{x}{1}\) gives tan \(\theta=\frac{x}{\sqrt{1-x^2}}\)
In this case also, tan (sin^-1x) = \(\frac{x}{\sqrt{1-x^2}}\) ........(3)
Equations (1), (2) and (3) establish that tan (sin⁻¹x) \(=\frac{x}{\sqrt{1-x^{2}}},-1