a = 2b {a=4, b=2}

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

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
\(\therefore\)  \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } =-{ cosec }^{ 2 }x\)

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

\({ sin }^{ -1 }\left[ \frac { x+\sqrt { 1-{ x }^{ 2 } } }{ \sqrt { 2 } } \right] \quad Let\quad x=sin\theta \Rightarrow \theta ={ sin }^{ -1 }x\)
\(={ sin }^{ -1 }\left( \frac { sin\quad \theta +\sqrt { 1-{ sin }^{ 2 } } \theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( \frac { sin\theta +cos\theta }{ \sqrt { 2 } } \right) \)
\(={ sin }^{ -1 }\left( sin\theta \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } cos\theta \right) \)
\(={ sin }^{ -1 }\left( sin\theta cos\frac { \pi }{ 4 } +cos\theta sin\frac { \pi }{ 4 } \right) \)
\(={ sin }^{ -1 }\left[ \theta +\frac { \pi }{ 4 } \right] \)
\(\Rightarrow \theta +\frac { \pi }{ 4 } =\frac { \pi }{ 4 } +{ sin }^{ -1 }x\)

\({ tan }^{ -1 }\left[ \frac { \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } }\quad }{ \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } } } \right] \)
\(\begin{cases} \because 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 },0\le x\le \frac { \pi }{ 2 } \\ and\quad 1-sin\quad x={ \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } +\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } }{ \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } -\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } +sin\frac { x }{ 2 } +cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } -cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( cot\frac { x }{ 2 } \right) \)
\(={ tan }^{ -1 }tan\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) =\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) \)

Given A = Set of all human beings in a town at a particular time and R = {(x, y); x is a wife of y; x, y \(\in \) A} 
(i) Since x is a wife of x, is not true (x, x) \(\notin \)  R
So, R is not reflexive. 
(ii) x is a wife of y, but y not wife of (y, x) \(\notin \) : R. So, R is not symmetric. 
(iii) x is a wife of y, and y is wife of z But this situation does not exist. So, R is not transitive.

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

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\)  A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

We have, \(\left[ \begin{matrix} 2x+y & x-y \\ x-z & x+y+z \end{matrix} \right] =\left[ \begin{matrix} 10 & -1 \\ 2 & 8 \end{matrix} \right] \)
\(\Rightarrow\) 2x + y = 10, x - y = - 1
x - z = 2 and x + y + z = 8
\(\therefore\) 2(y - 1) + y = 10  \(\Rightarrow\)  2y + y + 2 = 10
\(\Rightarrow\)3y = 12  \(\Rightarrow\)  y = 4
\(\therefore\)  x = 3
3 - z,  \(\Rightarrow\)  z = 1
\(\therefore\)  x = 3, y = 4, z = 1

\(2-20=2x^{ 2 }-24 \)
\(\Rightarrow x=\pm \sqrt { 3 }\)

\(=\frac { 1 }{ log\quad a } .\frac { 1 }{ sin\quad x } .cos\ x=\frac { cot\quad x }{ log\quad a } \)

\(We\quad have:\)
\(X=\sqrt { { a }^{ { sin }^{ -1 }t } } and\quad y=\sqrt { { a }^{ { cos }^{ -1 }t } } \)
\(i.e.,x={ \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }and\quad y={ \left( { \alpha }^{ { cos }^{ -1 }t } \right) }^{ \frac { 1 }{ 2 } }\)
\(\frac { dx }{ dt } =\frac { 1 }{ 2 } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }^{ -\frac { 1 }{ 2 } }\frac { d }{ dt } { \left( { \alpha }^{ { sin }^{ -1 }t } \right) }\)
\(=\frac { 1 }{ 2 } \frac { 1 }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } { \alpha }^{ { sin }^{ -1 }t }{ log }_{ e }\alpha \frac { d }{ dx } \left( { sin }^{ -1 }t \right) \)
\(=\frac { { log }_{ e }\alpha }{ 2 } .\sqrt { { \alpha }^{ { sin }^{ -1 }t } } .\frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(Similarly\quad \frac { dy }{ dt } =-\frac { { log }_{ e }\alpha }{ 2 } \sqrt { { \alpha }^{ { cos }^{ -1 }t } } \frac { 1 }{ \sqrt { 1-{ t }^{ 2 } } } \)
\(\frac { dy }{ dx } =\frac { dy/dt }{ dx/dt } =\frac { \sqrt { { \alpha }^{ { cos }^{ -1 }t } } }{ \sqrt { { \alpha }^{ { sin }^{ -1 }t } } } =-\frac { y }{ x } .,\quad which\quad is\quad true.\)

We shall prove the result by using principle of mathematical induction
\(\mathrm{P}(n): \text { If } \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^n=\left[\begin{array}{cc} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right], n \in \mathbf{N}\)
\(P(1): A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, so } A^1=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\)
Therefore, the result is true for n = 1.
Let the result be true for n = k. So
\(\mathrm{P}(k): \mathrm{A}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \text {, then } \mathrm{A}^k=\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right]\)
Now, we prove that the result holds for n = k +1
\( \mathrm{A}^{k+1} =\mathrm{A} \cdot \mathrm{A}^k=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{array}\right] \)
\(=\left[\begin{array}{cc} \cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\ -\sin \theta \cos k \theta+\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta \end{array}\right]\)
\(=\left[\begin{array}{cc} \cos (\theta+k \theta) & \sin (\theta+k \theta) \\ -\sin (\theta+k \theta) & \cos (\theta+k \theta) \end{array}\right]=\left[\begin{array}{cc} \cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta \end{array}\right]\)
Therefore, the result is true for n = k + 1. Thus by principle of mathematical induction, we have \(A^n=\left[\begin{array}{rr} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{array}\right]\), holds for all natural numbers.

\(We\quad know\quad that\quad A={ I }_{ 2 }A\)
\(i.e.\ \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}A\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 6 & -3 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}A\quad [Applying\quad { R }_{ 1 }\leftrightarrow { R }_{ 2 }]\)
\( \Rightarrow \begin{bmatrix} -2 & 1 \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 1 & 3 \end{bmatrix}A\quad [Applying\quad { R }_{ 2 }\rightarrow { R }_{ 2 }+3{ R }_{ 1 }]\)
Since second row of LHS matrix has all zero,
\(\therefore \ { A }^{ -1 }\) does not exist.

\(Here\quad y={ sec }^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) \)
\(={ sec }^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) \)
\(={ sec }^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) \left( \frac { 1 }{ cos2\theta } \right)\)
\( { sec }^{ -1 }(sec\quad 2\theta )=2\theta =2{ cos }^{ -1 }x\)
\(Hence\quad \frac { dy }{ dx } =-\frac { 2 }{ \sqrt { 1-{ x }^{ 2 } } } \)

(i) 'f' is a function.
[∵ each element of A in the first place in the ordered pair is related to only one element of A in the second place]
(ii) 'g' is a not function.
[∵ 1 is related to two elements of A namely 2 and 3

LHS:\(\begin{vmatrix} 2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}\)
= \(\begin{vmatrix}x+y+z&x+y+z&x+y+z\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix}1&1&1\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix} 1&0&0\\2z&0&-(x+y+z)\\x-y-z&x+y+z&0\end{vmatrix}\)
= \((x+y+z)(1)\begin{vmatrix}0&-(x+y+z)\\x+y+z&0 \end{vmatrix}\)
= \((x+y+z)[0+(x+y+z)^2]\)
= \((x+y+z)^3\)= RHS

The given system of equation is:
2x - y = -2
3x + 4y = 3
these can be written as A X = B
= X = A^-1B
where  \(A=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}, X=\begin{bmatrix} x\\y\end{bmatrix}\ and\ B=\begin{bmatrix}-2\\3 \end{bmatrix}\)
\(|A|=\begin{bmatrix} 2&-1\\3&4\end{bmatrix}=8+3=11\neq0\Rightarrow A^{-1}\)exists.
Now \(adj\ A=\begin{bmatrix} 4&-3\\1&2\end{bmatrix}=\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\)
\(A^{-1}={1\over |A|}(adj\ A)={1\over11}\begin{bmatrix}4&1\\3&2 \end{bmatrix}\)
\(X={1\over11}\begin{bmatrix}4&1\\-3&2 \end{bmatrix}\begin{bmatrix} -2\\3\end{bmatrix}\)
\(={1\over11}\begin{bmatrix} -8+.3\\6+6\end{bmatrix}={1\over11}\begin{bmatrix} -5\\12\end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} -5/11\\12/11\end{bmatrix}\)
\(x={-5\over11},y={12\over11}\)|

Let's consider \(\tan ^{-1}(1)=x\). Then, \(\tan x=1=\tan \left(\frac{\pi}{4}\right)\). \(\therefore \tan ^{-1}(1)=\frac{\pi}{4}\)
Let's assume,\(\cos ^{-1}\left(-\frac{1}{2}\right)=y\).
Then, \(\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)\)
\(\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)
Let's again assume that \(\sin ^{-1}\left(-\frac{1}{2}\right)=z\).
Then, \(\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\).
\(\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)
\(\therefore \tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
\(=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \)
\(=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

Consider \(f: \{ 1,2,3,4\} \rightarrow \{ 1,2,3,4\} \)
and  \(g:\{ 1,2,3,4\} \rightarrow \{ 1,2,3\} \) defined by:
\(f(1)=1,f(2)=2,f(3)=f(4)=3\)
\(g(1)=1,g(2)=2,g(3)=g(4)=3\).
It can be seen that gof is onto but f is not onto.

\(LHS={ tan }^{ -1 }\frac { 1 }{ 2 } +{ tan }^{ -1 }\frac { 2 }{ 11 } \)
\(={ tan }^{ -1 }\frac { \frac { 1 }{ 2 } +\frac { 2 }{ 11 } }{ 1-\frac { 1 }{ 2 } .\frac { 2 }{ 11 } } ={ tan }^{ -1 }\frac { 15 }{ 20 } ={ tan }^{ -1 }\frac { 3 }{ 4 } =RHS.\)

\(\text { We have } \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}\)
\(\text { Or }\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4}\)
\(\text { i.e. }\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\frac{\pi}{4}\)
\(\text { Therefore } \quad \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4}=1\)
\(\text { or } \quad 6 x^{2}+5 x-1=0 \text { i.e., }(6 x-1)(x+1)=0\)
\(\text { which gives } \ x=\frac{1}{6} \text { or } x=-1 .\)
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

The binary operation * on N is defined as a * b = L.C.M. of a and b.
(i) 5 * 7 = L.C.M. of 5 and 7 = 35
20 * 16 = L.C.M of 20 and 16 = 80
(ii) It is known that:
L.C.M of a and b = L.C.M of b and a &mn For E; a, b ∈ N.
∴ a * b = b * a
Thus, the operation * is commutative.
(iii) For a, b, c ∈ N, we have:
(a * b) * c = (L.C.M of a and b) * c = LCM of a, b, and c
a * (b * c) = a * (LCM of b and c) = L.C.M of a, b, and c
∴ (a * b) * c = a * (b * c)
Thus, the operation * is associative.
(iv) It is known that:
L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N
⇒ a * 1 = a = 1 * a &mnForE; a ∈ N
Thus, 1 is the identity of * in N.
(v) An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a.
Here, e = 1
This means that:
L.C.M of a and b = 1 = L.C.M of b and a
This case is possible only when a and b are equal to 1.
Thus, 1 is the only invertible element of N with respect to the operation *.

Given \(asec^{ 3 }\theta \) 
\(\therefore \frac { dx }{ d\theta } =3asec^{ 3 }\theta tan\theta \)
\(y=atan^{ 3 }\theta \)
\(\therefore \frac { dx }{ d\theta } =tan^{ 2 }\theta sec^{ 2 }\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 3atan^{ 2 }\theta sec^{ 2 }\theta }{ 3asec^{ 3 }\theta tan\theta } =sin\theta \)
\(\frac { dy }{ dx } =sin\frac { \pi }{ 4 } \)
\( =\frac { 1 }{ \sqrt { 2 } } \)

\(A=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] \)
then \(adj\quad A={ \left[ \begin{matrix} 4 & -3 \\ -1 & -2 \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
Taking \(A(adj\quad A)=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] { \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Taking \((adj\ A)A={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }{ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Getting \(\left| A \right| =\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] =-11\)
\(\therefore \left| A \right| =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
A.(adj A) = (adj A).A = |A| Hence proved.

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \ -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.

Given * on Q, defined by a * b = ab + 1
Let, a \(\in Q\), b E Q then
ab \(\in Q\),
and (ab + 1)  \(\in Q\),
\(\Rightarrow \) a * b = ab + 1is defined on Q
" is a binary operation on Q.
Commutative:
a*b = ab + 1
= ab + 1 
= a*b = b*a
So * is commutative on Q.
Associative:
(a "b) * c = (ab + 1) * c = (ab + l)c + 1
= abc + c + 1
a " (b " c) = a " (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
So * is associative on Q. 1
Identity Element: Let e \(\in Q\) be the identity element,
then for every a 
a * e = a and e * a = a
ac + 1 = a and ca + 1 = a
\(\Rightarrow e=\frac { a-1 }{ a } ,\frac { a-1 }{ a } \)
e is not unique as it depend on 'a', hence identity element does not exist for *.
Inverse: since there is not identity element, hence there is no inverse.

\(L.H.S.={ tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 1 }{ 2 } +\frac { 1 }{ 5 } }{ 1-\frac { 1 }{ 2 } \times \frac { 1 }{ 5 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 7 }{ 10 } }{ \frac { 9 }{ 10 } } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) \)
\(={ tan }^{ -1 }\left( \frac { 7 }{ 9 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 8 } \right) ={ tan }^{ -1 }\left( \frac { \frac { 7 }{ 9 } +\frac { 1 }{ 8 } }{ 1-\frac { 7 }{ 9 } \times \frac { 1 }{ 8 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \frac { 65 }{ 72 } }{ \frac { 65 }{ 72 } } \right) \)
\(={ tan }^{ -1 }(1)\)
\(=\frac { \pi }{ 4 } =R.H.S.\)

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

Let Rs. 30,000 be divided into two parts:
Rs. x invested in 1st type and Rs. (30,000-x) in 2^nd type.
The values of the bonds are represented by \(1\times 2\) row matrix as: \(A=\left[ \begin{matrix} x & 30,000-x \end{matrix} \right] \)
The amount received as interest per Rs. annually are represented by \(2\times 1\) column matrix as: \(B=\left[ \begin{matrix} \frac { 5 }{ 100 } \\ \frac { 7 }{ 100 } \end{matrix} \right] .\)
\(\therefore \) The interest to be obtained is a \(1\times 1\) matrix, which is represented by the product matrix \(1\times 1\).
\(AB=\left[ \begin{matrix} x & 30,000-x \end{matrix} \right] \left[ \begin{matrix} \frac { 5 }{ 100 } \\ \frac { 7 }{ 100 } \end{matrix} \right] \)
\(=\left[ \frac { 5x }{ 100 } +\left( 30,000-x \right) \frac { 7 }{ 100 } \right] \)
\(=\left[ 2100-\frac { 2x }{ 100 } \right] .\)
(a) By the question,
\(\left[ 2100-\frac { 2x }{ 100 } \right] =\left[ 1800 \right] \)
\(\Rightarrow 2100-\frac { 2x }{ 100 } =1800\)
\(\Rightarrow \frac { 2x }{ 100 } =300\)
\(\Rightarrow x=15000\)
Hence, the reqd.amounts are Rs.15,000 and Rs. (30,000 - 15,000) i.e., Rs. 15,000 and Rs. 15,000.
(b) By the question,
\(\left[ 2100-\frac { 2x }{ 100 } \right] =\left[ 2000 \right] \)
\(\Rightarrow 2100-\frac { 2x }{ 100 } =2000\)
\(\Rightarrow \frac { 2x }{ 100 } =100\)
\(\Rightarrow x=5000\)
Hence, the reqd .amounts are Rs. 5,000 and Rs. (30,000 - 5,000)
i.e., Rs. 5,000 and Rs. 25,000.          

As differentiable functions is continuous also

As cos(-680°) = cos 680°
= cos(720° – 40°) = cos 40°
∴ cos^-1[cos(-680°)J = cos^-1 (cos 40°)
= 40° = \(\frac{2\pi}{9}\).