\(\frac { { a }^{ 2 }sin(B-C) }{ sinA } =\frac { { (2RsinA) }^{ 2 }sin(B-C) }{ sinA } =\frac { { 4R }^{ 2 }{ sin }^{ 2 }Asin(B-C) }{ sinA } \)
= 4R²sin A sin(B- C) = 4R²sin(B + C) sin (B - C)
= 4R²(sin²B - sin²C) = 4R²sin²B - 4R²sin²C
= b²- c²
Similarly, \(\frac { { b }^{ 2 }sin(C-A) }{ sinB } ={ c }^{ 2 }-{ a }^{ 2 }\)
\(\frac { { c }^{ 2 }sin(A-B) }{ sinC } ={ a }^{ 2 }-{ b }^{ 2 }\)
\(\frac { { a }^{ 2 }sin(B-C) }{ sinA } +\frac { { b }^{ 2 }sin(C-A) }{ sinB } +\frac { { c }^{ 2 }sin(A-B) }{ sinC } ={ b }^{ 2 }-{ c }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 }=0\) 

By sine formula \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =2R\)
\(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { c }^{ 2 } } =\frac { { (2RsinA) }^{ 2 }-{ (2RsinB) }^{ 2 } }{ { (2RsinC) }^{ 2 } } \)
\(=\frac { 4{ R }^{ 2 }{ sin }^{ 2 }A-4{ R }^{ 2 }{ sin }^{ 2 }B }{ 4{ R }^{ 2 }{ sin }^{ 2 }C } =\frac { { sin }^{ 2 }A-{ sin }^{ 2 }B }{ { sin }^{ 2 }C } \)
\(=\frac { sin(A+B)sin(A-B) }{ { sin }^{ 2 }C } [\because sinC=sin(A+B)]\)
\(=\frac { sin(A+B)sin(A-B) }{ { sin }^{ 2 }(A+B) } =\frac { sin(A-B) }{ sin(A+B) } \) 

For 4 digit numbers, we have to arrange the given 5 digits in 4 vacant places. This can be done in ⁵p₄ = 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 = 120 ways.

For 4-digit even numbers, the unit's place can be occupied by one of the 2 digits 2 or 4. The remaining 4 digits can be arranged in the remaining 3 places in ⁴p₃ ways.
\(\therefore\) By the multiplication rule, the required number of 4-digit even numbers is 2 x ⁴P₃ = 2 \(\times\) 4 \(\times\) 3 \(\times\) 2 = 48.

Here ^{n - 1}P₃ : ⁿP₄ = 1 : 9
\(\therefore {(n-1)!\over(n-4)!}:{n!\over(n-4)!}=1:9\)
\(\Rightarrow {(n-1)!\over(n-4)!}\times{(n-4)!\over n!}={1\over9} \Rightarrow {(n-1)!\over n!}={1\over9} \Rightarrow {(n-1)!\over n(n-1)!}={1\over9}\)
\(\Rightarrow {1\over n}={1\over9}\Rightarrow n=9\)

Here, \(\frac{S_{3}}{S_{6}}=\frac{125}{152}\)
⇒ \(\frac{a(r^{3}-1)/(r-1)}{a(r^{6}-1)/(r-1)}=\frac{125}{152}\Rightarrow \frac{r^{3}-1}{r^{6}-1}=\frac{125}{152}\)
∴ \(\frac{r^{3}-1}{(r^{3}-1)(r^{3}+1)}=\frac{125}{152}\Rightarrow \frac{1}{r^{3}+1}=\frac{125}{152}\)
∴ 152 = 125 r³ + 125 or 125r³ = 27
ஃ r3=\(\frac{27}{125}=(\frac{3}{5})^{3}\)
⇒ r = {\((\frac{3}{5})^{3}\)}^1/3 = \(\frac{3}{5}\)
Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

\(sec(\alpha+\frac{\pi}{2})=-cosec\alpha\)
\(LHS=[(1+cot\alpha)+cosec\alpha][(1+cot\alpha)-cosec\alpha]\)
\(=(1+cot\alpha)^2-cosec^2\alpha\)
\(=1+cot^2\alpha+2cot\alpha-cosec^2\alpha \ [\because 1+cot^2\alpha=cosec^2\alpha]\)
\(=cosec^2\alpha+2cot\alpha-cosec^2\alpha\)
\(=2cot\alpha=RHS\)

Let A be the first term and D, the common difference of A.P.
a_p = a, ∴ A + (p -1)D = a --- (1)
aq = b, ∴ A+(q-1)D = b --- (2)
a, = c, ∴ A+(r-1)D = c --- (3)
∴ a (q - r) + b (r - p) + c (p - q) = [A + (p - 1) D] (q - r) + [A + (q -1) D]
(r - p) + [A + (r - 1) D] (p - q) [Using (1), (2) and (3)]
= (q - r + r - p + P - q) A + [ (p -1)(q - r) + (q -1)(r - p) + (r -1) (p - q)] D
= (0) A + (pq - pr - q + r + qr - pq - r + p + pr - p - qr + q) D
= (0) A + (0) D =0.

Slope of the line joining the points (2, 3) and (3, -1) is
\(\frac { -1-3 }{ 3-2 } \)=-4
Slope of the required line which is perpendicular to it
=\(\frac { -1 }{ -4 } =\frac { 1 }{ 4 } \)   [∵ m₁m₂=-1]
Equation of the line passing through the point (5, 2) is
y-2 =\(\frac { 1 }{ 4 } \)(x-5)    [y-y₁=m(x-x₁)]
⇒ 4y-8=x-5
⇒ x-4y+3=0

Let T_r+1 be the term in whichx³² and x^-17 occurs,
\(\therefore T_{r+1}= {^{15}C_{r}}.(x^{4})^{15-r}(-\frac{1}{x^{3}})^{r}\)
= \({^{15}C_{r}},(-1)^{r},x^{60-4r},x^{-3r}={^{15}C_r},(-1)^{r}, x^{60-7r}\)
(i) Since x³² occurs in this term
∴ Exponent of x = 32
⇒ 60 - 7r = 32 ⇒ 7r = 28
∴ r = 28 ÷ 7 = 4
∴ Coefficient ofthe term containing x³² is = ¹⁵C₄(-1)⁴ = 1365
(ii) Since x^-17occurs in this term
∴ Exponent of x = -17
⇒ 60-7r = -17
⇒ 7r = 77,  ∴ r = 11
∴ Coeffiicciient of the term containing x^-17=¹⁵C₁₁(-1)¹¹= -¹⁵C₁₁(-1)¹¹ = -¹⁵C_15-11 = -¹⁵C₄= -1365.

zero

Since a, b, c are in A.P
\(b=\left( \frac { a+c }{ 2 } \right) \)....(1)
and b, c, d are in G.P\(\Rightarrow\) c = \(\sqrt { bd } \)......(2)
Also c,d,e in H.P \(\Rightarrow\) d\(\frac { 2ce }{ c+e } \)
From (2), C² = bd = \(\left( \frac { a+c }{ 2 } \right) \left( \frac { 2ce }{ c+e } \right) \)
\(=\left( \frac { (a+c)ce }{ c+e } \right) \)
\(\Rightarrow\)  \(c=\frac { (a+c)e }{ c+e } \)
\(\Rightarrow\) C²+ c e = ae + ce
\(\Rightarrow\) c² = ae
a, c, e are in G.P

Let the A,P be a, a+d, a + 2d,...
Given S_n = 3n²
We have T_n = n-S_n-1
⇒ T_n = 3n²-3(n-1)²
⇒ T_n = 3n²-3(n²-2n + 1)
⇒ T_n = 3n²- 3n² + 6n - 3
⇒ T_n = 6n - 3
∴   T₁ = 6(1)-3 = 3
T₂ = 6(2)-3 = 9
T₃ = 6(3)-3 = 15 
∴ The A.P is 3, 9, 15...

Given p(n): 3ⁿ>n. to P.T. p(n+1) is true
i.e to P.T. 3ⁿ⁺¹ > n+1
Since p(n) is true, 3ⁿ>n
\(\Rightarrow\) 3.3ⁿ > 3n     [multiply by 3 both sides]
\(\Rightarrow\) 3ⁿ⁺¹>n+2n
\(\Rightarrow\) 3ⁿ⁺¹ > n+1   \(\left[ \because \quad 2n>1\ for\ every\ n\in N\Rightarrow 2n+n>n+1\ for\ every\ n\in N \right] \)
\(\Rightarrow\) p(n+1) is true.

Here T₂-T₁ = \(log\frac { { a }^{ 2 } }{ b } -log\quad a=log\frac { { a }^{ 2 }/b }{ a } \)
= \(log\frac { a }{ b } \)
T₃-T₂ = \(log\frac { a^{ 3 } }{ b^{ 2 } } -log\frac { a^{ 2 } }{ b^{ 2 } } =log\frac { a^{ 3 } }{ b^{ 2 } } +log\frac { a^{ 2 } }{ b^{ 2 } } \)
= \(log\frac { a^{ 3 } }{ b^{ 2 } } \times \frac { b }{ { a }^{ 2 } } =log\frac { a }{ b } \)
∴ T₂-T₁ = T₃-T₂ = \(log\left( \frac { a }{ b } \right) \)
∴ The given sequences is an A.P

Using binomial theorem, we have
(1 + ax)ⁿ-1 + nC₁(ax) + nC₂(ax)²+.......+aⁿxⁿ
= \(1+nax+{{n(n-1)}\over{2}}a^2x^2+....a^nx^n\)
Given (1 + ax)ⁿ = 1 + 12x + 64x² +....
Conparing the Co-efficient of x and x², we get
n a = 12
and \({n(n-1)\over 2}a^2=64\)
\((n-1).{na.a\over2}=64\Rightarrow(n-1){(12)a\over2}=64\)
\((n-1)6a=64\Rightarrow(n-1)a={{64}\over{6}}\)    \(\left[ \because na=12\Rightarrow a={12\over n} \right]\)
\(\Rightarrow(n-1)\left( {12\over n} \right)={64 \over 6}\)
\(={n-1\over n}={ 64 \over 6\times 12}\Rightarrow{n-1\over n}={8\over 9}\)
\(\Rightarrow\)  9n - 9 = 8n
\(\Rightarrow\)  n = 9 and \(a=\frac { 12 }{ n } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)

There are 10 questions in Part-A, out of which 8 questions can be chosen in 10C₈ ways.
Similarly from Part-B, contains using 10 questions, 5 questions can be chosen in 10C₅ ways.
Hence, the total number of ways of selecting 8 questions from Part-A and 5 from Part-B.
= 10C₈ \(\times\)10C₅ = \(\frac { 10! }{ 8!2! } \times \frac { 10! }{ 5!\times 5! } \)

= 11340.

The number of arrangements in which the best and the worst papers never come together can be obtained by subtracting from the total number of arrangements, the number of arrangements in which the best and worst come together.
Number of arrangements of 9 papers = 9P₉ = 9!
Considering the best and the worst paper as one paper, we have 8 papers which can be arranged in 8P₈ = 8! ways.
But the best and worst papers can be put together in 2! ways.
So, the number of permutations in which the best and worst papers can be put together = 2! \(\times\) 8! ways.
Hence, the number of ways in which the best and the worst papers never come together.
= 9! - 2! 8!
=9 \(\times\) 8! - 2! 8! =8! (9 - 2)   [\(\because\) 9! = 9\(\times\)8!]
= 8! \(\times\) 7 = 8 \(\times\) 7 \(\times\) 6 \(\times\) 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 \(\times\) 7
= 282240

Given (n+2)! = 60(n-1)!

\(\Rightarrow \) (n+2) (n+1)(n) = 60
\(\Rightarrow \) (n+2)(n+1)(n) = 5 \(\times\)4 \(\times\)3
Equating the terms both sides we get, n = 0

LHS = \(cos\left( \frac { \pi }{ 4 } -A \right) cos\left( \frac { \pi }{ 4 } -B \right) -sin\left( \frac { \pi }{ 4 } -A \right) sin\left( \frac { \pi }{ 4 } -B \right) \)
= \(cos\left\{ \left( \frac { \pi }{ 4 } -A \right) \left( \frac { \pi }{ 4 } -B \right) \right\} \)    [cos A cos B - sin A sin B = cos(A + B)
= \(cos\left( \frac { \pi }{ 2 } -A-B \right) \)    where A = \(\frac { \pi }{ 4 } -A,B=\frac { \pi }{ 4 } -B\)
= \(cos\left( \frac { \pi }{ 2 } -(A+B \right) \)
sin(A + B) = RHS

Given R is the set of all real numbers. Then R x R is the set of all ordered pairs (x, y) where x, y ∈ R.
R \(\times\)R = {(x,y): x, y ∈ R}
Clearly R \(\times\)R is the set of all points in xy-plane. Now R \(\times\)R \(\times\)R = {(x, y, z) : x,y, x ∈ R}.
∴ R \(\times\)R \(\times\)R represents the set of all points in space.

The venn diagram of A and B are disjoint sets but both are subsets of C is

The shaded region is A' U (A ∩ B) or (A\B)'

\({x^2+1\over x(x+1)^2}={A\over x}+{B\over x+1}+{c\over x(x+1)^2}\)  [Since the denominator is linear factor of repeated factors]
\(⇒\ {x^2+1\over x(x+1)^2}={A(x+)^2+Bx(x+1)+Cx\over x(x+1)^2}\)
⇒ x²+ 1 = A(x + 1)² + Bx(x + 1) + Cx
Putting x = 0, we get
1 = A ⇒ = 1
Putting x = -1, we get
1 + 1 = C(-1) ⇒ 2 = -C ⇒ C = -2
Putting x = 1 in (1) we get
2 = A(4) + B(2) + C
⇒ 24A + 2B + C   [∴ A = 1, C = -2]
⇒ 2 = 4 + 2B - 2
⇒ 2 + 2 - 4 = 2B ⇒ B =0
\(∴\ {x^2+1\over x(x+1)^2}={1\over x}+{0\over x+1}-{2\over (x+1)^2}\)
\(\frac { { x }^{ 2 }+1 }{ x\left( x+1 \right) ^{ 2 } } =\frac { 1 }{ x } \frac { 2 }{ \left( x+1 \right) ^{ 2 } } \)

Given quadratic equation is
5^2x-5^x+3 + 125 = 5x
\(\Rightarrow\) (5^x)²-5^x 5³- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-125 5^x- 5^x + 125 = 0
\(\Rightarrow\) (5^x)²-126.5^x+125=0
Let 5^x=y
\(\Rightarrow\)y²-126y + 125 = 0
\(\Rightarrow\)(y-1)(y-125) = 0
\(\Rightarrow\)    y = 1 or 125
\(\Rightarrow\)   5^x = 1 or 125
Case (i) When 5^x = 1 \(\Rightarrow\) 5⁰ \(\Rightarrow\) x = 0
Case (ii) When 5^x = 125 \(\Rightarrow\) 5^x = 5³\(\Rightarrow\) x = 3
\(\therefore\) The roots are 0, 3.

Given x > 2
Converting the into equation, we get x = 2. 
Clearly it is a line parallel to y-axis at a distance of 2 units from it.
Also, we find that the point (0, 0) does not satisfy the inequation x > 2
So, the region represented by the given equation is the shaded region.