\(sin^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] =\frac { 2\pi }{ 5 } \)
Alternative Method :
\(sin^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] =sin^{ -1 }\left[ sin\left( \pi -\frac { 3\pi }{ 5 } \right) \right] \)
\(\left( \because \frac { 3\pi }{ 5 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right) \)
\(sin^{ -1 }\left( sin\frac { 2\pi }{ 5 } \right) =\frac { 2\pi }{ 5 } \)

\(Let\quad { sec }^{ -1 }x=\theta ,\quad then\quad x=sec\theta \quad and\quad for\quad x<-1,\)
\(\frac { \pi }{ 2 } <\theta <\pi \)
Given expression = cot^-1(-cot \(\theta \))
\(={ cot }^{ -1 }\left[ cot\left( \pi -\theta \right) \right] =\pi -{ sec }^{ -1 }x\quad as\quad 0<\pi -\theta <\frac { \pi }{ 2 } \)

\(R.H.S.=2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } \quad \quad \begin{cases} Let\quad x=cos\theta \\ \Rightarrow \theta ={ cos }^{ -1 }x \end{cases}\)
\(=2{ sin }^{ -1 }\left( \sqrt { \frac { 1-cos\theta }{ 2 } } \right) \)
\( \begin{cases} as\quad cos\theta =1=2{ sin }^{ 2 }\frac { \theta }{ 2 } \\ \Rightarrow 1-cos\theta =2sin^{ 2 }\frac { \theta }{ 2 } \end{cases}\)
\(=2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
\(=2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }x\)
L.H.S = R.H.S

\(L.H.S.={ tan }^{ -1 }\left( \frac { 3{ a }^{ 2 }x-{ x }^{ 3 } }{ { a }^{ 3 }-3a{ x }^{ 2 } } \right) \)
\(Divide\quad by\quad a^{ 3 },\)
\(={ tan }^{ -1 }\left( \frac { 3\frac { x }{ a } -\frac { { x }^{ 3 } }{ { a }^{ 3 } } }{ 1-3\left( \frac { { x }^{ 2 } }{ { a }^{ 2 } } \right) } \right) \)
\(={ tan }^{ -1 }\left[ \frac { 3\left( \frac { x }{ a } \right) -{ \left( \frac { x }{ a } \right) }^{ 3 } }{ 1-3{ \left( \frac { x }{ a } \right) }^{ 2 } } \right] \)
\(Put,\frac { x }{ a } =tan\theta ,\quad \theta ={ tan }^{ -1 }\frac { x }{ a } \)
\(={ tan }^{ -1 }\left[ \frac { 3tan\theta -{ tan }^{ 3 }\theta }{ 1-3{ tan }^{ 2 }\theta } \right] \)
\(={ tan }^{ -1 }(tan\quad 3\theta )\)
\(3\theta =3{ tan }^{ -1 }\left( \frac { x }{ a } \right) =R.H.S.\)

\({ tan }^{ -1 }\frac { 3 }{ 4 } +{ tan }^{ -1 }\frac { 3 }{ 5 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(={ tan }^{ -1 }\left( \frac { \frac { 15+12 }{ 20 } }{ \frac { 20-9 }{ 20 } } \right) -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x+y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\frac { 27 }{ 11 } -{ tan }^{ -1 }\frac { 8 }{ 19 } \)
\(\left[ \because { tan }^{ -1 }x-{ tan }^{ -1 }y={ tan }^{ -1 }\left( \frac { x-y }{ 1+xy } \right) \right] \)
\(={ tan }^{ -1 }\left( \frac { \frac { 27 }{ 11 } -\frac { 8 }{ 19 } }{ 1+\frac { 27\times 8 }{ 11\times 19 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 513-88 }{ 209 } }{ \frac { 209+216 }{ 209 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { 425 }{ 425 } \right) \)
\(={ tan }^{ -1 }1=\frac { \pi }{ 4 } \)

\({ tan }^{ -1 }\left[ \frac { cos\quad x }{ 1+sin\quad s } \right] \) \(\quad \because \) \(\begin{cases} cos\quad x={ cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } \\ and\quad 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { { cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] Divide\quad by\quad cos\frac { x }{ 2 } ,\quad we\quad get\)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { x }{ 2 } }{ 1+tan\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) \right] =\frac { \pi }{ 4 } -\frac { x }{ 2 } \)

\(\cot \left(\tan ^{-1} a+\cos ^{-1}\right) \)
\(=\cot \left(\frac{\pi}{2}\right) \quad\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \)
\(=0\)

\(\left[\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\right]=\frac{2 \pi}{3}\)
Alternative Method:
\( {\left[\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\right]}\)
\( =\left[\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right]\)
\( =\left[\frac{\pi}{3}+2 \times \frac{\pi}{6}\right]\)
\( {\left[\because \quad \cos ^{-1}(\cos \quad \theta)=\theta \forall \theta[0, \quad \pi] \quad \text { and } \sin ^{-1}(\sin \quad \theta)=\theta \forall \theta=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]} \)
\( =\frac{\pi}{3}+\frac{\pi}{3} =\frac{2 \pi}{3}\)

\({ sin }^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] =\frac { 2\pi }{ 5 } \)
Alternative Method :
\({ sin }^{ -1 }\left[ sin\left( \frac { 3\pi }{ 5 } \right) \right] ={ sin }^{ -1 }\left[ sin\left( \pi -\frac { 3\pi }{ 5 } \right) \right] \)
\(\left( \because \quad \frac { 3\pi }{ 5 } \notin \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \right) \)
\({ sin }^{ -1 }\left( sin\frac { 2\pi }{ 5 } \right) =\frac { 2\pi }{ 5 } \)

\( y=\operatorname{cosec}^{-1}(-\sqrt{2}) \)
\(\Rightarrow \operatorname{cosec} y=-\sqrt{2} \)
\(\Rightarrow \operatorname{cosec} y=\operatorname{cosec}\left(-\frac{\pi}{4}\right) \)
We know that the range of the principal value branch of cosec⁻¹x is \( {\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} .} \)
\(\therefore y=-\frac{\pi}{4} \)
Hence, the principal value of  \(\operatorname{cosec}^{-1}(-\sqrt{2}) \text { is }-\frac{\pi}{4}\)

Let \({ cosec }^{ -1 }\left( -\sqrt { 2 } \right) =y\), where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] -\left\{ 0 \right\} \)
\(\Rightarrow cosecy=-\sqrt { 2 } =-cosec\frac { \pi }{ 4 } \)
\(=cosec\left( -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 4 } \)
Hence, the required principal value = \(-\frac { \pi }{ 4 } \)

Let tan⁻¹(−√3) = θ ⇒ tan θ = −√3
We know that the range of principal value of tan⁻¹ θ is (−π/2,π/2).
∴ tanθ = −√3=−tan π/3=tan(−π/3) (∵ tan(−θ) = −tanθ)
⇒ θ =−π/3, where θ ϵ (−π/2,π/2) ⇒tan⁻¹(−√3)=−π/3
Hence, principal value of tan−1(−√3) is −π/3

\(\text {Let } \sin ^{-1} \frac{12}{13}=x, \quad \cos ^{-1} \frac{4}{5}=y, \tan ^{-1} \frac{63}{16}=z\)
\( \sin x=\frac{12}{13}, \quad \cos y=\frac{4}{5}, \quad \tan z=\frac{63}{16}\)
\(\cos x=\frac{5}{13}, \sin y=\frac{3}{5}, \tan x=\frac{12}{5} \text { and } \tan y=\frac{3}{4}\)
\(\text {We have }\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5} \times \frac{3}{4}}=-\frac{63}{16}\)
Hence tan(x + y) = − tan z
i.e., tan (x + y) = tan (–z) or tan (x + y) = tan (p – z)
Therefore x + y = – z or x + y = p – z
Since x, y and z are positive, x + y \(\ne\) – z (Why?)
\(\text {Hence }x+y+z=\pi \text { or } \sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi\)

Take x = cos θ, then proceeding as above, we get, \({ \sin }^{ -1 }(2x\sqrt { 1-{ x }^{ 2 } } )\) = 2 cos^–1 x

Let \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) =y\), Then \(\sin y=\frac{1}{\sqrt{2}}\)
We know that the range of the principal value branch of \(\sin ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\)
Therefore, principal value of \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) is \ \frac { \pi }{ 4 } \)

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

sin^-1(1-x) - 2sin x \(=\frac { 1 }{ 2 } \)
sin^-1(1 - x) = (1 + 2sin^-1x)
(1 - x) = sin \(\left( \frac { 1 }{ 2 } +2sin^{ -1 }x \right) \)
sin^-1x = Q
x = sinQ
Cos2 Q = 1 - 2sin²Q
cos2Q = 1 - 2x²
\(\Rightarrow \) 1 - x = cos2Q
1 - x = 1 - 2x²
\(\Rightarrow \) 2x² - x = 0
x (2x - 1) = 0
x = 0 or x = \(\frac { 1 }{ 2 } \)
Put \(\frac { 1 }{ 2 } \)in equation
sin \(\frac { 1 }{ 2 } \) 2sin^-1 \(\frac { 1 }{ 2 } \)
\(=\frac { n }{ E } \times -2\times \frac { n }{ 6 } \)
So x = 0

Given that sin[ cot ^-1(x + 1)] = cos(tan^-1x)....(i)
We know that,
\({ cot }^{ -1 }(A)={ sin }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x + 1
Applying this identity in equation (i), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos({ tan }^{ -1 }x)\) ...(ii)
Also, we know that
\({ tan }^{ -1 }A={ cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ A }^{ 2 } } } \)
Here, A = x
Applying this identity in equation (ii), we have
\(sin\left[ { sin }^{ -1 }\frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } \right] =cos\left( { cos }^{ -1 }\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \right) \)
\(\Rightarrow \frac { 1 }{ \sqrt { 1+(1+{ x) }^{ 2 } } } =\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)
\(\left[ \because { sin }^{ -1 }(sin\theta )=\theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
and cos^-1 (cos \(\theta \)) = \(\theta \) \(\forall \theta \) [0, \(\pi \)]]
Squaring and Reciprocating both side, we have
1+(1 + x)² = 1 + x²
\(\Rightarrow\) 1 + 1 + x² + 2x = 1 + x²
\(\Rightarrow\) 1 + 2x = 0
\(\Rightarrow\)  \(x=-\frac { 1 }{ 2 } \quad \)

\(y=\cot ^{ -1 }{ (\sqrt { \cos { x } } ) } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } ,\)
\(\Rightarrow y=\frac { \pi }{ 2 } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } -\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } \)
\(\Rightarrow y=\frac { \pi }{ 2 } -2\tan ^{ -1 }{ (\sqrt { \cos { x } } ) } \)
\(\Rightarrow y=\frac { \pi }{ 2 } -\left[ \frac { (2\sqrt { \cos { x } } ) }{ 1-{ (\sqrt { \cos { x } } ) }^{ 2 } } \right] \)
\(\Rightarrow \left[ \frac { (2\sqrt { \cos { x } } ) }{ 1-{ (\sqrt { \cos { x } } ) }^{ 2 } } \right] =\frac { \pi }{ 2 } -y\)
\(\Rightarrow \frac { (2\sqrt { \cos { x } } ) }{ 1-\cos { x } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ \sqrt { { (2\sqrt { \cos { x } } ) }^{ 2 } } +{ (1+\cos { x } ) }^{ 2 } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ \sqrt { 1+\cos ^{ 2 }{ x } +2\cos { x } } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ { \sqrt { (1+\cos { x } } ) }^{ 2 } } \)
\(\Rightarrow \sin { y } =\frac { 1-\cos { x } }{ 1+\cos { x } } =\frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } \)
\(\Rightarrow \sin { y } =\tan ^{ 2 }{ \frac { x }{ 2 } } \)