On differentiating e^y (x+1) = 1
e^y+(x+1)e^ydy/dx = 0
\(\Rightarrow { e }^{ y }+\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } ={ -e }^{ y }\)

We have y = log(sin x)
dy/dx = d/dx \(\left| log(sinx) \right| \)
= \(\frac { 1 }{ sinx } \times cosx\)
\(\Rightarrow\) dy/dx = cot x
\(\therefore\)  \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } =-{ cosec }^{ 2 }x\)

Let f : [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers, then there exists some c in (a, b) such that f^'(c) = 0.

(i) f(g) = f(x)g(x)
\(fg=(\sqrt { x+4 } )(\sqrt { x-4 } )=\sqrt { x^{ 2 }-4 } \)
(ii) \(\frac { f }{ g } =\frac { f(x) }{ g(x) } =\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x+4 } }{ \sqrt { x-4 } } \times \frac { \sqrt { x-4 } }{ \sqrt { x-4 } } \)
\(=\frac { \sqrt { x^{ 2 }-16 } }{ x-4 } \)

Let x = cos 2\(\theta \)
\(=sin\left[ 2t{ an }^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \right] \)
\(=sin\left[ 2tan^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \right] \)
\(\left[ \because cos2\theta =1-2{ sin }^{ 2 }\theta \ and\ cos\ 2\theta =2{ cos }^{ 2 }\theta -1 \right] \)
\(=sin\left[ 2{ tan }^{ -1 }\left( tan\quad \theta \right) \right] \)
\(=sin(2\theta )=\sqrt { 1-{ cos }^{ 2 }2\theta } \)
\(=sin\quad 2\theta =\sqrt { 1-{ x }^{ 2 } } \)

We have \(x\in [-1,\quad 1]\)
Let  x = sin \(\theta\)             \(\therefore \quad \theta \in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow \quad \quad \theta ={ sin }^{ -1 }x\)
\(\Rightarrow \quad \quad -\frac { \pi }{ 2 } \le \theta \le \frac { \pi }{ 2 } \)
\(\Rightarrow \quad \quad \frac { \pi }{ 2 } \ge -\theta \ge -\frac { \pi }{ 2 } \)
\(\Rightarrow \quad \frac { \pi }{ 2 } +\frac { \pi }{ 2 } \ge \frac { \pi }{ 2 } -\theta \ge -\frac { \pi }{ 2 } +\frac { \pi }{ 2 } \)
\(\Rightarrow \quad\pi \ge \frac { \pi }{ 2 } -\theta \ge 0\)
\(\Rightarrow \quad cos\left( \frac { \pi }{ 2 } -\theta \right) =sin\quad \theta =x\)
\(\Rightarrow \quad \frac { \pi }{ 2 } -\theta ={ cos }^{ -1 }x\)
\(\Rightarrow { \quad cos }^{ -1 }x=\frac { \pi }{ 2 } -{ sin }^{ -1 }x\)
\(\Rightarrow { \quad sin }^{ -1 }x+{ cos }^{ -1 }x=\frac { \pi }{ 2 } if\quad x\in [-1,\quad 1]\)

\(f(x)=x^{ 2 },x\in Ra\)
\(f(1.1)=(1.1)^{ 2 }z\)
\(=1.21\)
\(f(1)=(1)^{ 2 }=1\)
\(\frac { f(1.1)-f(1) }{ 1.1-1 } =\frac { 1.21 }{ 1.1-1 } =\frac { 0.21 }{ 0.1 } \)
\(=2.1\)

Let A be a skew-symmetric matrix. Then by definition \({ A }^{ \prime }=-A\)
\(\Rightarrow\) the (i, j)^th element of \({ A }^{ \prime }\) = the (i, j)^th element of (- A)
\(\Rightarrow\) the (j, i)^th element of A = - the (i, j)^th element of A
For the diagonal elements i = j \(\Rightarrow\) the (i, j)the element of A = - the (i, j)^th element of A.
\(\Rightarrow\)  the (i, j)^th element of A = 0
Hence the diagonal elements are all zero.

We have, \(X+Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] ,X-Y=\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(\left( X+Y \right) +\left( X-Y \right) =\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] +\left[ \begin{matrix} 6 & 5 \\ 7 & 3 \end{matrix} \right] \)
\(2X=\left[ \begin{matrix} 8 & 8 \\ 12 & 4 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(\therefore \ X=\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
and \(Y=\left[ \begin{matrix} 2 & 3 \\ 5 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 4 \\ 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -2 & -1 \\ -1 & -1 \end{matrix} \right] \)

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) =\frac { \pi }{ 6 } \)
Alternative Method :
\({ tan }^{ -1 }\left( tan\frac { 7\pi }{ 6 } \right) ={ tan }^{ -1 }\left[ tan\left( \pi +\frac { \pi }{ 6 } \right) \right] \)
\(={ tan }^{ -1 }\left[ tan\frac { \pi }{ 6 } \right] =\frac { \pi }{ 6 } \)
\(\left[ \because { tan }^{ -1 }(tan\theta )=\theta \forall \theta \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)

a = 1
Alternate Method :
Given \(\left[ \begin{matrix} a-b & 2a+c \\ 2a-b & 3c+d \end{matrix} \right] =\left[ \begin{matrix} -1 & 5 \\ 0 & 13 \end{matrix} \right] \)
By equating,   a - b = - 1.. (i)
and  2a - b = 0... (ii)
By (ii) - (i),
2a - b - a + b = 0 + 1
⇒ a = 1

\(|A|=6-5=1, \text { adj } A=\left[\begin{array}{rr} 3 & -5 \\ -1 & 2 \end{array}\right] \)
\(\therefore \ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \)
\(\Rightarrow A^{-1}=\frac{1}{1}\left[\begin{array}{rr} 3 & -5 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{rr} 3 & -5 \\ -1 & 2 \end{array}\right] \)

(i) 2 * 3 = 2³ = 8
(ii) 3 * 2 = 3² = 9.

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

\(\text { We have } \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\)
\(\text { Or }\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}\)
\(\text { i.e. }\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\text { Therefore } \quad \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}=1\)
\(\text { or } \quad 6 x^{2}+5 x-1=0 \text { i.e., }(6 x-1)(x+1)=0\)
\(\text { which gives } \ x=\frac{1}{6} \text { or } x=-1 .\)
Since x = – 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative\(x=\frac{1}{6}\) is the only solution of the given equation

Given A = \(\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right] \)
and B = \(\left[ \begin{matrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{matrix} \right] \)
\(\therefore AB=\left[ \begin{matrix} 2+4 & 2-2 & -4+4 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ -4+4 & 2-2 & -4+10 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix} \right] \)
\(\therefore \) AB = 6I
Premultiplying by A^-1
A^-1AB = 6A^-1I
=> IB = 6A^-1I  (\(\therefore \) A^-1A = I)
=> B = 6A^-1   (\(\therefore \) IX = X)
=> A^-1 = 1/6B
Given equations are:
x - y = 3
2x + 3y + 4z = 17
y + 2z = 7
=> \(\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] \)
AX = C, where
C = \(\left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] \)
X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
=> A^-1AX = A^-1C
=> X = A^-1C
=> X = \(\frac { 1 }{ 6 } \left[ \begin{matrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 17 \\ 7 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 12 \\ -6 \\ 24 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 4 \end{matrix} \right] \)
\(\therefore \) x = 2, y = -1 and z = 4

Hence, the given function is not differentiable at x=2.

We have, \(\begin{equation} A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right] \end{equation}\) 
For applying row operations, write A as A = lA, 
where I is an identity matrix of order 3.
i.e \(\begin{equation} \left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \end{equation}\) 
Here, a₁₁ = 0, so on applying R₁ ↔️ R₂' we get
Applying R₃➝R₃ we get
\(\begin{equation} \left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{array}\right]=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{array}\right] \end{equation}\) 
Applying R_1➝ R_{l ➝} 2R₂ and R_{3 ➝} R₃ + 5R₂, we get
\(\begin{equation} \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array}\right]=\left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{array}\right] \end{equation}\) 
Applying \(\begin{equation} R_{3} \rightarrow \frac{1}{2} R_{3} \end{equation}\) we get
\(\begin{equation} \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{rcc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] A \end{equation}\) 
Applying R₁ ➝ R_l + R₃ and R₂ ➝ R₂ - 2R₃,we get
\(\begin{equation} \left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2 \end{array}\right] \end{equation}\) 
Now, new matrix equation is in the form of
I = BA , where \(\begin{equation} B=\left[\begin{array}{ccc} 1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2 \end{array}\right] \end{equation}\) 
Hence,
\(\begin{equation} A^{-1}=\left[\begin{array}{ccc} 1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2 \end{array}\right] \end{equation}\)

Given equation can be written as
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] \) or AX = B
|A| = 1(7)+1(19)+2(-11)
= 4\(\neq \)0
\(\therefore \)A^-1 exists
Co-factors
a₁₁ = 7, a₁₂ = -19 a₁₃ = -11
a₂₁ = 1, a₂₂ = -1  a₂₃ = -1
a₃₁ = -3, a₃₂ = 11 a₃₃ = 7
\(\Rightarrow { A }^{ -1 }=\frac { 1 }{ 4 } \left[ \begin{matrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{matrix} \right] \)
\(\therefore \) X = A^-1B
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{matrix} \right] \left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore \) x = 2, y = 1 and z = 3

\(\text {We have } 2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sec x) \)
\(\Rightarrow \tan ^{-1}\left(\frac{2 \sin x}{1-\sin ^{2} x}\right)=\tan ^{-1}\left(\frac{2}{\cos x}\right) \)
\(\Rightarrow \frac{2 \sin x}{\cos ^{2} x}=\frac{2}{\cos x} \Rightarrow \frac{\sin x}{\cos x}=1 \)
\(\Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4} \)

Clearly the function is defined at every real number. By inspection, it seems prudent to partition the domain of definition of f into three disjoint subsets of the real line.
Let \( \mathrm{D}_1=\{x \in \mathbf{R}: x<0\}, \mathrm{D}_2=\{0\} \text { and } \mathrm{D}_3=\{x \in \mathbf{R}: x>0\} \)
Case 1 At any point in D1 , we have f(x) = x 2 and it is easy to see that it is continuous there
Case 2 At any point in D3 , we have f(x) = x and it is easy to see that it is continuous there
Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f(0) = 0. The left hand limit of f at 0 is
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x^2=0^2=0\)
The right hand limit of f at 0 is
\(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x=0\)
Thus \(\lim _{x \rightarrow 0} f(x)=0=f(0)\)and hence f is continuous at 0. This means that f is continuous at every point in its domain and hence, f is a continuous function

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\(\Rightarrow \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}A \quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}A \quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }]\)
\(\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}A.\quad [Applying\quad { R }_{ 1 }\rightarrow { R }_{ 1 }-{ 2R }_{ 2 }]\)
\(Hence,\quad { A }^{ -1 }=\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}.\)

(C) is the correct answer.
Reason.  \({ A }^{ 2 }=AA\)
\(=\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \gamma & -\alpha \end{bmatrix}\)
\(=\begin{bmatrix} { \alpha }^{ 2 }+\beta \gamma & \alpha \beta -\beta \alpha \\ \gamma \alpha -\alpha \gamma & \beta \gamma +{ \alpha }^{ 2 } \end{bmatrix}\)
\(=\begin{bmatrix} { \alpha }^{ 2 }+\beta \gamma & 0 \\ 0 & { \alpha }^{ 2 }+\beta \gamma \end{bmatrix}.\)
Now  \({ A }^{ 2 }=I\)
\(\Rightarrow { \alpha }^{ 2 }+\beta \gamma =1\)
\(\Rightarrow 1-{ \alpha }^{ 2 }-\beta \gamma =0.\)

2+4R₁ and R₃-->R₃+7R₁]
\(\Rightarrow 2[5(2+7sin3\theta)-10(cos2\theta+4sin3\theta)]=0\) [Expanding by C₂]
\(\Rightarrow 10[(2+7sin3\theta)-(2cos2\theta+8sin\theta)]=0\)
\(\Rightarrow 2+7sin3\theta-2cos2\theta-8sin3\theta=0\)
\(\Rightarrow 2-2cos2\theta-sin3\theta=0\)
\(\Rightarrow 2-2(1-2sin^2\theta)-(3sin\theta-4sin^3\theta)=0\)
\(\Rightarrow 4sin^3\theta-3sin\theta+4sin^2\theta=0\)
\(\Rightarrow sin\theta(4sin^2\theta+4sin\theta-3)=0\)
\(\Rightarrow sin\theta(2sin\theta-1)(2sin\theta+3)=0\)
\(\Rightarrow sin\theta=0\ or\ sin\theta={1\over2}\), which is true
[∵ sinθ=-\(3\over2\) is not true]

Since a * b = L.C.M of a,b,
\(\therefore \) 2*3 = L.C.M of 2 and 3 = 6 \(\notin \){1, 2, 3, 4, 5}.
Hence ' * ' is not a binary operation.

Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x.
Then (h o g) (x) = h(g (x)) = h (1– x + | x |)
= |1– x + | x || = f(x)
we have seen that h is a continuous function.
Hence g being a sum of a polynomial function and the modulus function is continuous.
But then f being a composite of two continuous functions is continuous.

Let A=\(\left|\begin{matrix}1&2\\ 3&4\end{matrix}\right|\)
Then =\(A_11=(-1)^{1+1}M_{11}=(+1)(4)\)
\(=4\)
\(=A_{12}=(-1)^{1+2}M_{12}=(-1)(3)\)
\(=-3\)
\(A_{21}=(-1)^{2+1}M_{21}=(-1)(2)\)
\(=-2\)
\(A_{22}=(-1)^{2+2}M_{22}=(+1)(1)\)
\( =1\)
\(\therefore\ adj\ A=\begin{vmatrix} A_{11}&A_{12}\\A_{21}&A_{22}\end{vmatrix}'\)
\(\left[\begin{matrix}4&-3\\ -2&1\end{matrix}\right]'=\left[\begin{matrix}4&-2\\ -3&1\end{matrix}\right]\)

Let y be an arbitrary element of range f. Then y = 4x² + 12x + 15, for some
x in N, which implies that y = (2x + 3)² + 6. This gives \(x=\frac{((\sqrt{y-6})-3)}{2}, \text { as } y \geq 6\)
\(\text { Let us define } g: \mathrm{S} \rightarrow \mathbf{N} \text { by } g(y)=\frac{((\sqrt{y-6})-3)}{2}\)
Now gof (x) = g(f (x)) = g(4x² + 12x + 15) = g((2x + 3)² + 6)
\(=\frac{\left(\left(\sqrt{(2 x+3)^{2}+6-6}\right)-3\right)}{2}=\frac{(2 x+3-3)}{2}=x\)
\(\text { and }\operatorname{fog}(y)=f\left(\frac{((\sqrt{y-6})-3)}{2}\right)=\left(\frac{2((\sqrt{y-6})-3)}{2}+3\right)^{2}+6\)
\(=((\sqrt{y-6})-3+3))^{2}+6=(\sqrt{y-6})^{2}+6=y-6+6=y\)
Hence, gof = I_N and fog =I_S. This implies that f is invertible with f ^–1 = g.

Let \({ sin }^{ -1 }\left( -\frac { 1 }{ 2 } \right) =y\) where \(y\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)
\(\Rightarrow siny=-\frac { \pi }{ 2 } \)
\(\Rightarrow siny=-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right) \)
\(\Rightarrow y=-\frac { \pi }{ 6 } \)
Hence, the required principal value = \(-\frac { \pi }{ 6 } \)

Let x = sec θ, then \(\sqrt{x^2-1}=\sqrt{\sec ^2 \theta-1}=\tan \theta\)
Therefore, \(\cot ^{-1} \frac{1}{\sqrt{x^2-1}}=\cot ^{-1}(\cot \theta)=\theta=\sec ^{-1} x\) which is the simplest form

As x^o = \(\frac { \pi }{ 180 } { x }^{ c }\)
\(\therefore \frac { d }{ dx } (cot{ x }^{ o })=\)\(\frac { d }{ dx } \left( cot\frac { \pi }{ 180 } x \right) \)
\(=-\frac { \pi }{ 180 } { cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }\frac { \pi }{ 180 } x\)
\(=-{ 1 }^{ o }{ cosec }^{ 2 }{ x }^{ 0 }\)

As \(\begin{bmatrix} 5 & x \\ y & 0 \end{bmatrix}=\begin{bmatrix} 5 & x \\ y & 0 \end{bmatrix}\Rightarrow x=y\)

As sin^-1 x = \(\frac { \pi }{ 2 } \), sin^-1 y = \(\frac { \pi }{ 2 } \), sin^-1z = \(\frac { \pi }{ 2 } \)
⇒ x = 1, y = 1, z = 1
∴ x + y² + z³ = 1 + 1 + 1 = 3