\(|A|=\left| \begin{matrix}1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right| =1+{ tan }^{ 2 }x={ sec }^{ 2 }x\neq 0\)
\(\Rightarrow\) A^-1 exists
Let C_ij be the cofactor of a_ij  in A
C₁₁ = (-1)¹⁺¹ M₁₁ = (-1)²(1) = 1
C₁₂ = (-1)¹⁺² (-tan x) = tan x
C₂₁ = (-1)²⁺²(1) = 1
\(\therefore \quad adj\quad A={ \left[ \begin{matrix} 1 & tan\quad x \\ -tan\quad x & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1+{ tan }^{ 2 }x } \left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(\therefore \quad { A }^{ T }{ A }^{ -1 }=\left[ \begin{matrix} 1 & -tan\quad x \\ tan\quad x & 1 \end{matrix} \right] \left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \frac { -tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { tan\quad x }{ 1+{ tan }^{ 2 }x } +\frac { tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \frac { 1 }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] \)
\(=\left[ \begin{matrix} \frac { 1-tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { -2tan\quad x }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tan\quad x }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } \end{matrix} \right] =\left[ \begin{matrix} cos\quad 2x & -sin2x \\ sin\quad 2x & cos\quad 2x \end{matrix} \right] \)  (Using multiple angle formula)

Give A = \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0\)
Applying R₂ \(\rightarrow\) R₂-R₁ and R₃ \(\rightarrow\) R₃ - R₁ 
we get A = \(\left| \begin{matrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-a & c-b \end{matrix} \right| =0\)
Expanding along C₁ we get
\(1\left| \begin{matrix} c-a & a-b \\ b-a & c-b \end{matrix} \right| =0\)
\(\Rightarrow\) (c - a) (c - b) - (b - a) (a - b) = 0
\(\Rightarrow\) c² - bc - ac + ab - (ab - b² - a² + ab) = 0
\(\Rightarrow\) a² + b² + c² - ab - bc - ca = 0
Multiplying both sides by 2 we get, 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
\(\Rightarrow\) (a - b)² + (b - c)² + (c - a)² = 0
\(\Rightarrow\) a = b = 0, b - c = 0,c - a = 0
\(\Rightarrow\) a = b = c
\(\Rightarrow\) \(\Delta\) ABC is equilateral.
\(\therefore A=B=C=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ 2 }A+{ sin }^{ 2 }B+{ sin }^{ 2 }C=3{ sin }^{ 2 }\frac { \pi }{ 3 } =3{ \left( sin\frac { \pi }{ 3 } \right) }^{ 2 }=3{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=3\times \frac { 3 }{ 4 } =\frac { 9 }{ 4 } \)

Let A = \(\left| \begin{matrix} 2y+4 & \quad 5y+7 & 8y+a \\ 3y+5 & 6y+8 & 9y+b \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| \)
Applying R₂\(\rightarrow\) 2R₂ and dividing by 2 we get,
A = \(\frac { 1 }{ 2 } \left| \begin{matrix} 2y+4 & 5y+7 & 8y+a \\ 6y+10 & 12y+16 & 18y+2b \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| \)
Applying R₂\(\rightarrow\) 2R₂-(R₁ + R₃) we get
\(A=\frac { 1 }{ 2 } \left[ \begin{matrix} 2y+4 & \quad 5y+7 & 8y+a \\ 0 & 0 & 2b-(a+c) \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right] \)
Given a,b,c in A.P \(\Rightarrow\) 2b = a + c \(\Rightarrow\) 2b - (a + c) = 0
\(\therefore A=\frac { 1 }{ 2 } \left| \begin{matrix} 2y+4 & 5y+7 & 8y+a \\ 0 & 0 & 0 \\ 4y+6 & 7y+9 & 10y+c \end{matrix} \right| =0\)
\(\therefore\) |A| = 0

Given \(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42 & 40 & 2 \end{matrix} \right| =0\)
Applying C₁\(\rightarrow\)C₁ - C₂, we get,
\(\Delta =\left| \begin{matrix} { cosec }^{ 2 }\theta -{ cot }^{ 2 }\theta & { cot }^{ 2 }\theta & 1 \\ { cot }^{ 2 }\theta -{ cosec }^{ 2 }\theta & { cosec }^{ 2 }\theta & -1 \\ 42-40 & 40 & 2 \end{matrix} \right| \)
\(=\left| \begin{matrix} 1 & { cot }^{ 2 }\theta & 1 \\ -1 & { cosec }^{ 2 }\theta & -1 \\ 2 & 40 & 2 \end{matrix} \right| \)  [\(\because\) cosec²  \(\theta\)  - cot²  \(\theta\)  =1]   
= 0
\(\Delta =0\)    [\(\because\)  C₁ \(\equiv \) C₃]  

Given matrix A is singular, if |A| = 0
\(|A|=\begin{vmatrix} x+1&-3&4\\-5&x+2&2\\4&1&x-6 \end{vmatrix}=0\)
Expanding along R₁ we get,
\(|A|=x+1\begin{vmatrix}x+2 & 2 \\ 1 & x-6 \end{vmatrix}+3\begin{vmatrix} -5 & 2 \\ 4 & x-6 \end{vmatrix}+4\begin{vmatrix}-5 & x+2 \\ 4 & 1\end{vmatrix}=0\)
\(\Rightarrow\) (x + 1)[(x + 2)(x - 6) - 2] + 3[-5 (x - 6) - 8] + 4 [-5 - 4 (x + 2)] = 0
\(\Rightarrow\) (x + 1) [x² - 4x - 12 - 2] + 3[-5x + 30 - 8] + 4 [-5 - 4x - 8] = -0
\(\Rightarrow\) (x + 1)(x² - 4x - 14) + 3(-5x + 22) + 4(-4x - 13) = 0
\(\Rightarrow\) x³ - 4xl - 14x + xl - 4x - 14 - 15x + 66 - 16x - 52 = 0
\(\Rightarrow\) x³ - 3x² - 49x = 0
\(\Rightarrow\) x(x² - 3x - 49) = 0
\(\Rightarrow\)    \(x=0\ or\ x={{3\pm\sqrt{{(-3)}^{2}}-4(1)(-49)}\over{2a}}\)                                                           
\(\begin{bmatrix} \because\ x = {-b \pm \sqrt{b^2-4ac} \over 2a},a = 1, b = -3, c=-49 \end{bmatrix}\)
\(\Rightarrow\) \(x=0\ or\ x={{3\pm\sqrt{9+196}}\over{2}}\)
\(\Rightarrow\) \(x=0\ or\ x={{3\pm\sqrt{205}}\over{2}}\)