\(\text {LHS }=2 \sin ^2 \frac{3 \pi}{4}+2 \cos ^2 \frac{\pi}{4}+4 \sec ^2 \frac{3 \pi}{4}\)
\(=2 \sin ^2 135^{\circ}+2 \cos ^2 45^{\circ}+4 \sec ^2 135^{\circ}\)
\(=2 \sin ^2\left(180^{\circ}-45^{\circ}\right)+2\left(\frac{1}{\sqrt{2}}\right)^2+4 \sec ^2\left(180^{\circ}-45^{\circ}\right)\)
\(=2 \sin ^2 45^{\circ}+1+4 \sec ^2 45^{\circ}\)
\(=2\left(\frac{1}{\sqrt{2}}\right)^2+1+4(\sqrt{2})^2\)
\(=1+1+8=10=\mathrm{RHS}\)
Hence proved.

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

\(\lim _{ x\rightarrow 2 }{ \frac { { x }^{ 3 }+2 }{ x+1 } } \)
\(\lim _{ x\rightarrow 2 }{ \frac { { x }^{ 3 }+2 }{ x+1 } } \)=\(\frac { 2^{ 3 }+2 }{ 2+1 } =\frac { 10 }{ 3 } \)

\(f(x)={ x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } } \)
\(f\left( \frac { 1 }{ x } \right) \)= \(=\frac { 1 }{ { x }^{ 3 } } -{ x }^{ 3 }\)
\(f(x)+f{ \left( \frac { 1 }{ x } \right) }={ x }^{ 3 }-\frac { 1 }{ x^{ 3 } } +\frac { 1 }{ x^{ 3 } } -{ x }^{ 3 }=0\).

Since the given line are concurrent
\(\left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0\)
\(\left| \begin{matrix} 3 & 4 & -13 \\ 2 & -7 & 1 \\ a & -1 & -14 \end{matrix} \right| =0\)
\(\Rightarrow \) 3 (98 +1) - 4 (-28 -a ) -13 (-2 +7a) = 0
\(\Rightarrow \) - 297 +112 +4a +26 -91a = 0
\(\Rightarrow \) 435 =87a
\(\Rightarrow \) a =\(\frac { 435 }{ 87 } \)
\(\Rightarrow \) a = 5

\(\left|\begin{array}{ccc} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{array}\right|=0\)
x(5x - 2x)-2(2x + x)-1 (4 + 5) = 0
3x² - 6x - 9 = 0
x² - 2x - 3 = 0 (Divided by 3)
(x - 3)(x + 1) = 0
\(x=+3,-1\)

\((x-2 y)^{13}\)
\(T_{r+1}=n C_r x^{n-r} a^r\)
\(n=13, r=4\)
\(t_{r+1}=13 C_r x^{13-r}(-2 y)^r\)
\(t_5=13 C_4 x^{13-4}(-2 y)^4\)
\(=\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} x^9(16) y^4\)
\(=11440 x^9 y^4\)

(2a - 3b)4
\( (x+a)^n=n C_0 x^n+n C_1 x^{n-1} a+n C_2 x^{n-2} a^2 +\ldots n C_{n-1} x a^{n-1}+n C_n a^n \)
\((2 a-3 b)^4= 4 C_0(2 a)^4-4 C_1(2 a)^3(3 b) +4 C_2(2 a)^2(3 b)^2 -4 C_3(2 a)(3 b)^3+4 C_4(3 b)^4\)
= 16a⁴ - 4 (8a³) (3b) + 6 (4a²) (9b²) - 4 (2a)(27b³) + 81b⁴
= 16a⁴ - 96 a³ b + 216 a² b² - 216 ab³ + 81b⁴

\(\frac{\pi}{8}=\frac{180^{\circ}}{8}=22 \frac{1}{2}^{\circ}=22^{\circ} 30^{\prime}\)

\(m=\frac{-a}{b}=\frac{-7}{5}\)

\(u=\cfrac { { x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(\cfrac { du }{ dx } =\cfrac { \left( 1+{ x }^{ 2 } \right) \left( 2x \right) -{ x }^{ 2 }(2x) }{ \left( 1+{ x }^{ 2 } \right) ^{ 2 } } \)
\(= \cfrac { 2x }{ \left( 1+{ x }^{ 2 } \right) ^{ 2 } } \) 
and \(v={ x }^{ 2 }\)
\( \therefore \cfrac { dy }{ dx } =2x\)
\(\cfrac { du }{ dv } =\cfrac { \left( \frac { du }{ dx } \right) }{ \left( \frac { dy }{ dx } \right) } \)
\(=\cfrac { \left| \frac { 2x }{ \left( 1+{ x }^{ 2 } \right) ^{ 2 }} \right| }{ 2x } \)
\(=\cfrac { 1 }{ \left( 1+{ x }^{ 2 } \right) ^{ 2 } } \)

\(\underset { x\rightarrow 2 }{ lim } \frac { { x }^{ 2 }-4x+6 }{ x+2 } =\cfrac { \underset { x-2 }{ lim } \left( { x }^{ 2 }+4x+6 \right) }{ \underset { x\rightarrow 2 }{ lim } \left( x+2 \right) } \)
\( =\cfrac { \left( 2 \right) ^{ 2 }-4\left( 2 \right) +6 }{ 2+2 } =\cfrac { 1 }{ 2 } \)

The letter of the word CHAT in alphabetical order are A, C, H, T.
(i) Number of words starting with A = 3! = 6 C begins
(ii) Number of words starting with CA are 2! = 2
Now the words CH begins
(iii) After that we get the word CHAT = 1
\(\therefore\) Rank of CHAT is = 6 + 2 + 1 = 9

\(\mathrm{RHS}=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
Let \(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x\)
\(=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)\)
\(=2 \theta=2 \tan ^{-1} x=\text { LHS }\)
Hence proved.

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

\(\text {LHS }=(\cos \alpha-\cos \beta)^2+(\sin \alpha-\sin \beta)^2\)
\(=\left(2 \sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta}{2}\right)^2+\left(2 \sin \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}\right)^2\)
\(=4 \sin ^2 \frac{\alpha-\beta}{2}\left[\sin ^2 \frac{\alpha+\beta}{2}+\cos ^2 \frac{\alpha+\beta}{2}\right]\)
\(=4 \sin ^2 \frac{\alpha-\beta}{2}=\text { RHS }\)
Hence proved.

Any number greater than a million will contain all the seven digits.
Now, we have to arrange these seven digits, out of which 2 occur twice, 3 occurs twice and the rest are distinct.
The number of such arrangements =\(\frac { 7! }{ 2!3! } =\frac { 7\times 6\times 5\times 4\times 3! }{ 2\times 3! } =420\)
These arrangements also include those numbers which contain 0 at the million's place.
Keeping 0 fixed at the million place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.
These 6 digits can be arranged in \(\frac { 6! }{ 2!3! } =\frac { 6\times 5\times 4\times 3! }{ 2\times 1\times 3! } =60\quad ways\)
Hence, the number of required numbers = 420 - 60 = 360.

\(\mathrm{LHS}= \tan (\pi+x) \cot (x-\pi)-\cos (2 \pi-x) \cos (2 \pi+x)-\tan (\pi+x) \cot (\pi-x) -\cos (2 \pi-x) \cos (2 \pi+x)\)
\(=-(\tan x)(-\cot x)-(\cos x)(\cos x)\)
\(=\tan \left(\frac{1}{\tan x}\right)-\cos ^2 x\)
\(=1-\cos ^{ 2 }{ x } =\sin ^{ 2 }{ x } \ \left[ \because \ 1-\cos ^{ 2 }{ x } =\sin ^{ 2 }{ x } \right] \)
= RHS  Hence Proved.

Let P (n) denote the statement 1³ + 2³ + 3³ + ....... + n³ = \(\frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } \)
Put n = 1
LHS = 1³ = 1
\(=\frac { 1^2(2)^2 }{ 4 } \Rightarrow 1\)
LHS = RHS
\(\therefore\) P (1) is true
Let us assume that P(k) is true
p(k) : 1³ + 2³ + ..... + k³ = \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } \)
To prove that P(k+1) IS true
p(k) : 1³ + 2³ + ..... + k³ + (k + 1)³
\(=P(k)+(k+1)^3\)
= \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } +(k+1)^{ 3 }\)
\(=\frac{k^2(k+1)^2+4(k+1)^3}{4}\)
\(=\frac{(k+1)^2\left(k^2+4(k+1)\right)}{4}\)
\(=\frac{(k+1)^2\left(k^2+4 k+4\right)}{4}=\frac{(k+1)^2(k+2)^2}{4}\)
∴ p(k + 1) is true if P(k) is true.
∴ p(n) is true for all \(n\in N\)

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)