LHS = \(2{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) +{ tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)
= \({ tan }^{ -1 }\left( \frac { \frac { 2 }{ 3 } +\cfrac { 2 }{ 3 } }{ 1-\left( \frac { 2 }{ 3 } \right) \left( \frac { 2 }{ 3 } \right) } \right) ={ tan }^{ -1 }\left( \frac { \frac { 4 }{ 3 } }{ \frac { 9-2 }{ 9 } } \right) \)
= \({ tan }^{ -1 }\left( \frac { 4 }{ 3 } \times \frac { 9 }{ 7 } \right) ={ tan }^{ -1 }\left( \frac { 12 }{ 7 } \right) \)
= RHS
Hence proved

Let \({ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) =\theta \Rightarrow \frac { 3 }{ 5 } =cos\theta \)
\(\therefore sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 25-9 }{ 25 } } \)
= \(\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

Let sin-1(-1) = y where \(\frac { -\pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
Then \(sin^{ -1 }(-1)=y\Rightarrow sin\quad y=-1\)
\(-1=sin\left( \frac { -\pi }{ 2 } \right) \Rightarrow y=\frac { -\pi }{ 2 } \) \(\left[ \because \frac { -\pi }{ 2 } \epsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\( \therefore\)  The principal value of \({ sin }^{ -1 }(-1)\ is \ \frac { -\pi }{ 2 } \)

Let y = sec^-1 (-2). Then, sec y = -2
By the definition, the range of the principal value branch of y = sec⁻¹x is [0, \(\pi\)]\{\({{\frac{\pi}{2}}}\)}
Let us find y in [0, \(\pi\)] - {\({{\frac{\pi}{2}}}\)} such that sec y = -2
But, sec y = −2 \(\Rightarrow\) cos y = -\(\frac{1}{2}\)
Now, cos y = -\(\frac { 1 }{ 2 } =-cos\frac { \pi }{ 3 } =cos\left( \pi -\frac { \pi }{ 3 } \right) =cos\frac { 2\pi }{ 3 } \). Therefore, y = \(\frac{2\pi}{3}\)
since \(\frac{2\pi}{3}\in[0,\pi]\)\{\({{\frac{\pi}{2}}}\)}, the principal value of sec^-1(-2) is \(\frac{2\pi}{3}\)

\(sin\left[ \frac { \pi }{ 3 } -si{ n }^{ -1 }\left( -\frac { 1 }{ 2 } \right) \right] =sin\left[ \frac { \pi }{ 3 } -\left( -\frac { \pi }{ 6 } \right) \right] =sin\left( \frac { \pi }{ 2 } \right) =1\)

Let cosec⁻¹(−1) = y. Then, cosec y = −1
Since the range of principal value branch of y = cosec^-1x is\(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0} and
cosec \((-\frac{\pi}{2})=-1,\) 
We have y = \(-\frac{\pi}{2}\).
Note that \(-\frac{\pi}{2}\in\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0}.
Thus, the principal value of cosec⁻¹(−1) is −\(\frac{\pi}{2}\)

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

cos^-1\([cos(-\frac{\pi}{6})]\neq \frac{\pi}{6}.\)
Since \(\frac { -\pi }{ 6 } \notin \left[ 0,\pi \right] \) which is the principal domain of cosine function. [\(\therefore \) cos -\(\theta\) = cos \(\theta\)]

Let cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

Let sin^-1 \(\left( -\frac { 1 }{ 2 } \right) \) = y. Then sin y = -\(\frac{1}{2}\)
The range of the principal value of sin^-1x is \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and hence, Let us find y \(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) Such that sin y = -\(\frac{1}{2}\). Clearly, y = -\(\frac{\pi}{6}\)
Thus, the principal value of sin^-1\(\left( -\frac { 1 }{ 2 } \right) \) is -\(\frac{\pi}{6}\). This corresponds to -30^o.

cos (tan^-1x) = \(sin\left( { cot }^{ -1 }\frac { 3 }{ 4 } \right) \)
\(\Rightarrow sin\left( { tan }^{ -1 }\frac { 4 }{ 3 } \right) =sin\left( { sin }^{ -1 }\frac { 4 }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } \right) \)
\(\left[ \because { tan }^{ -1 }x={ sin }^{ -1 }\left( \frac { x }{ \sqrt { 1+{ x }^{ 2 } } } \right) \right] \)
\(\Rightarrow sin\left( { sin }^{ -1 }\left( \frac { 4 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)
\(\Rightarrow cos\left( { tan }^{ -1 }x \right) =cos\left(- { tan }^{ -1 }x \right) =\frac { 4 }{ 5 } \)
\(\left[ \because cosx=cos(-x) \right] \)
\(\Rightarrow { tan }^{ -1 }x={- tan }^{ -1 }x={ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) \)
\(\Rightarrow \tan ^{-1} x=-\tan ^{-1} x=\tan ^{-1} \frac{3}{4}\)
\(\left[ \because { cos }^{ -1 }x={ tan }^{ -1 }\sqrt { \frac { 1-{ x }^{ 2 } }{ x } } ;{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) ={ tan }^{ -1 }\sqrt { \frac { 1-\frac { 16 }{ 25 } }{ \frac { 4 }{ 5 } } } \right] \)
\(\Rightarrow x=\frac { 3 }{ 4 } ,\frac { 3 }{ 4 } \) 

Let \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) =\theta \Rightarrow cos\theta =\frac { 4 }{ 5 } \)
\(\therefore tan\theta =\frac { opp }{ adj } =\frac { 3 }{ 4 } \)
\(\Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) \)
LHS = \({ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 3 }{ 4 } \right) +{ tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)
\(\left[ \because { tan }^{ -1 }x+tan^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1-xy } \right) \right] \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 3 }{ 4 } +\frac { 3 }{ 5 } }{ 1-\left( \frac { 3 }{ 4 } \right) \left( \frac { 3 }{ 5 } \right) } \right) \)
= \({ tan }^{ -1 }\left( \cfrac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) ={ tan }^{ -1 }\left( \frac { 27 }{ 20 } \times \frac { 20 }{ 11 } \right) \)
=  \({ tan }^{ -1 }\left( \frac { 27 }{ 11 } \right) =RHS\)
Hence proved

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

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


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

Given g(x) = sin^-1 x + cos^-1x
From the definition of sin^-1x.
\(-1\le x\le 1\) ...(1)
Also from the definition of cos^-1x
\(-1\le x\le 1\)   .........(2)
\(\therefore \) From (1) & (2),
Domain ofg(x) = [-1, 1] U [-1, 1]
= [-1, 1]
Hence the domain of g(x) is [-1, 1].

\(-5\pi \le x\le 5\pi \) and cos x = -1
cos x = -1
\(cosx=cos\pi \)  \([\because cos\theta =cos\alpha \Rightarrow 2n\pi +\alpha ,n\varepsilon Z]\)
\(x=2n\pi +\pi ,n\varepsilon Z\quad \)
= \((2n+1)\pi ,n\varepsilon Z\)
\(\Rightarrow But-5\pi \le x\le 5\pi \)
Putting,\(=0,\pm 1,\pm 2\) and -3 we get
\(\Rightarrow x=-5\pi ,-3\pi ,-\pi ,3\pi ,5\pi \)
\(x=(2n+1)\ {\pi},n=0,\pm 1,\pm 2 \) and -3

Let tan⁻¹(−1) = y. Then, tan y = -1 = -tan\(\frac{\pi}{4}=tan(-\frac{\pi}{4})\)
As -\(\frac{\pi}{4}\in (-\frac{\pi}{2},\frac{\pi}{2}), tan^-1(-1)=-\frac{\pi}{ 3}\)
Now, cos^-1\((\frac{1}{2})\) = y implies cos y = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
As \(\frac{\pi}{3}\)\(\in\)[0, \(\pi\)], cos^-1 \((\frac{1}{2})=\frac{\pi}{3}\)
Now, sin^-1\((-\frac{1}{2})\) = y implies sin y = -\(\frac{1}{2}\) = sin(-\(\frac{\pi}{3}\)).
As -\(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}], sin^-1(-\frac{1}{2})=-\frac{\pi}{6}\)
Therefore, tan⁻¹(−1)+cos^-1\((\frac{1}{2})+sin^-1(-\frac{1}{2})=-\frac{\pi}{4}+\frac{\pi}{3}-\frac{\pi}{6}=-\frac{\pi}{12}\)

\(2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =x\) and\({ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow cosx=\frac { 1 }{ 2 } \)
\(\Rightarrow cosx=cos\frac { \pi }{ 3 } \) \(\left[ \therefore \frac { \pi }{ 3 } \varepsilon \left[ 0,\pi \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) 
[\(\therefore\) Principal domain of sin is \(\therefore \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) and principal domain of cos is\(\left[ 0,\pi \right] \)
\(siny=\frac { 1 }{ 2 } \)
\(siny=sin\frac { \pi }{ 6 } \)  \(\left[ \because \frac { \pi }{ 6 } \varepsilon \left[ \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right] \)
\(\Rightarrow x=\frac { \pi }{ 3 } \) and \(y=\frac { \pi }{ 6 } \)
\(\therefore \quad 2{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \)
= \(2\left( \frac { \pi }{ 3 } \right) +\frac { \pi }{ 6 } =\frac { 2\pi }{ 3 } +\frac { \pi }{ 6 } \)
= \(\frac { 4\pi +\pi }{ 6 } =\frac { 5\pi }{ 6 } \)

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\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

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 \(\in[-b,-a]\cup[a,b]\).

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

Given \({ tan }^{ -1 }\left( \frac { \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } } \right) =a\)
\(\Rightarrow \frac { \left( \sqrt { 1+{ x }^{ 2 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) }{ \left( \sqrt { 1+{ x }^{ 3 } } -\sqrt { 1-{ x }^{ 2 } } \right) \left( \sqrt { 1+{ x }^{ 2 } } +\sqrt { 1-{ x }^{ 2 } } \right) } \)
= \(\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \frac { 2\sqrt { 1+{ x }^{ 2 } } }{ -2\sqrt { 1-{ x }^{ 2 } } } =\frac { tan\alpha +1 }{ tan\alpha -1 } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { 1-tan\alpha }{ 1+tan\alpha } \)
\(\Rightarrow \sqrt { \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } } =\frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \left( \frac { cos\alpha -sin\alpha }{ cos\alpha +sin\alpha } \right) ^{ 2 }\)
\(\Rightarrow \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } =\frac { 1-sin2\alpha }{ 1+sin2\alpha } \Rightarrow { x }^{ 2 }=sin2\alpha \)

Put \(x=a\ cos\theta \)
\(f(x)={ tan }^{ -1 }\sqrt { \frac { a-acos\theta }{ a+acos\theta } } ={ tan }^{ -1 }\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } } \)
= \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\frac { \theta }{ 2 } }{ 2{ cos }^{ 2 }\frac { \theta }{ 2 } } } =tan|tan\frac { \theta }{ 2 } |={ tan }^{ -1 }\left( tan\frac { \theta }{ 2 } \right) \)
= \(\frac { \theta }{ 2 } \) \([\because-a \(\left.\Rightarrow 0<\frac{\theta}{2}<\frac{\pi}{2} \therefore\left|\tan \frac{\theta}{2}\right|=\tan \frac{\theta}{2}\right]\)
= \(\frac { 1 }{ 2 } .{ cos }^{ -1 }\left( \frac { x }{ a } \right) \)\(\left[ \because x=acos\theta \Rightarrow cos\theta =\frac { x }{ a } \Rightarrow { cos }^{ -1 }\left( \frac { x }{ a } \right) \right] \)

The domain of sin^-1x is [-1, 1]
\(\therefore\) f(x) = sin^-1(2x - 3) is defined for all x, satisfying
\(-1\le 2x-3\le 1\)
\(\Rightarrow 3-1\le 2x\le 1+3\)
\(\Rightarrow 2\le 2x\le 4\Rightarrow 1\le x\le 2\Rightarrow x\epsilon \left[ 1,2 \right] \)
\(\therefore\) Domain of f(x) = sin^-1(2x - 3) is [1, 2].
(ii) The domain of f(x) is [-1, 1] and that of cosx is R
\(\therefore\) Domain of f(x) = sin^-1x + cos x is
\(\left[ -1,1 \right] \cap R=\left[ -1,1 \right] \)

Consider \({ tan }^{ -1 }\left( x-1 \right) +{ tan }^{ -1 }\left( x+1 \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { x-1+x+1 }{ 1-(x-1)(x+1) } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-\left( { x }^{ 2 }-1 \right) } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 }+1 } \right) \)
= \({ tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) \)
\(\therefore { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( x \right) ={ tan }^{ -1 }\left( 3x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( 3x \right) -{ tan }^{ -1 }\left( x \right) \)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 3xx }{ 1+3x^{ 2 } } \right) \)
\(\Rightarrow \frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 2x }{ 1+{ 3x }^{ 2 } } \)
x(1+ 3x²) = x(2-x²)
x[1+ 3x²-2 + x²] = 0
x(4x²-1) = 0
x(2x+ 1) (2x-1) = 0
x = 0, x = \(\frac{1}{2}\), x = -\(\frac{1}{2}\) are the roots 
Hence, there are 3 solutions

\({ sin }^{ -1 }\frac { 5 }{ x } +{ sin }^{ -1 }\frac { 12 }{ x } =\frac { \pi }{ 2 } \)
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