\(4\over 5\)

\(\theta=2n\pi-{\pi\over4}\)

From 5! onwards for all n! the unit digit is zero and hence the contribution to the unit digit is through 2! + 3! + 4! only. which is 2 + 6 + 24 = 32. Therefore the required unit digit is 2.

\(\frac { 6! }{ n! } =\frac { 1.2.3.4.5.6 }{ 1.2.3...n } \) = 6. As n < 6 we get, n = 5

When n = 50, r = 47
\(\frac { n! }{ r!(n-r)! } =\frac { 50! }{ 47!(50-47)! } =\frac { 50\times 49\times 48\times 47! }{ 47!\times 3! } =\frac { 50\times 49\times 48 }{ 1\times 2\times 3 } \) = 19600

A car can be bought by choosing a brand, then a variant model, and then a colour. A brand can be chosen in 2 ways; a model can be chosen in 3 ways and a colour can be chosen in 5 ways. By the rule of product, the person can buy a car in 2 \(\times\) 3 \(\times\)5 = 30 different ways.

(i) A student can choose a Mathematics book in "75" different ways.
(ii) A student can choose a Physics book in "35" different ways.
Hence applying the Rule of Sum, the number of ways a student can choose a book is 75 + 35 = 110.

LHS = sin²B + sin²C
= sin²B +\({ \left[ sin\left( \frac { \pi }{ 2 } -B \right) \right] }^{ 2 }\)= sin²B + cos²B
= 1 = RHS
Hence proved 

328⁰
328⁰ = 270⁰ + 58⁰
∴ 328⁰ lies in the IV quadrant

-55⁰
Since the given angle is negative, it moves in the clockwise direction
∴ -55⁰ lies in the IV quadrant

Let θ = 18°, Then 5θ = 90°
3θ + 2θ = 90° ⇒2θ = 90°-3θ
sin 2θ = sin (90° - 3θ) = cos 3θ
2 sin θ cos θ = 4 cos³θ - 3 cos θ. Since cos θ = cos 18° ≠ 0, we have
2 sin θ = 4 cos²θ - 3 = 4 (1 - sin²θ) - 3
4 sin²θ + 2 sin θ - 1 = 0
\(sin\ \theta=\frac{-2\pm\sqrt{4-4(4)(-1)}}{2(4)}=\frac{-1\pm\sqrt 5}{4}\)
Thus, \(\sin\ 18^o=\frac{\sqrt 5 -1}{4}\) (positive sign is taken sin 18^o is in I quadrant sinθ is positive).

(i) sin (A + B + C)
= sin A cos (B + C) + cos A sin (B + C)
= sin A cos B cos C + cos A sin B cos C + cos A cos B sin C - sin A sin B sin C
(ii) \(\tan (A+B+C) =\tan [A+(B+C)] \)
\(=\frac{\tan A+\tan (B+C)}{1-\tan A \tan (B+C)} \)
\(=\frac{\tan A+\frac{\tan B+\tan C}{1-\tan B \tan C}}{1-\tan A \frac{\tan B+\tan C}{1-\tan B \tan C}} \)
\(=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}\)

Let p(n) be the statement
\(Sin \alpha +sin\left(\alpha+{\pi \over6}\right)+...+sin\left(\alpha+(n-1){\pi\over 6}\right)={sin\left[\alpha+{(n-1}^\pi\over12\right]\times sin\left(n\pi\over 12\right)\over sin\left(\pi\over 12\right)}\)
Step 1: Putting n = 1, we get,
\(sin α={sin (α).sin{\pi\over12}\over sin{\pi\over 12}}=sin α\)
∴ p(1) is true
Step 2 : Let us assume that p(k) is true
\(sin α+sin\left(α+{\pi\over 6}\right)+...+sin \left(α+(k-1){\pi\over 6}\right)={sin\left(α+{(k-1)\pi\over12}\right)\times sin{k\pi\over 12}\over sin\left(\pi\over 12\right)}\)
Step 3 : To prove that p(k+1) is true
i.e. to p.T \(sin α+sin\left(α+{\pi\over 6}\right)+...+sin\left(α+(k-1){\pi\over 6}\right)+sin\left(α+{k\pi\over 6}\right)={sin\left(α+k{\pi\over 12}\right).sin{(k+1)\pi\over 12}\over sinm\left(\pi\over 12\right)}\)
LHS = \(sin α+sin\left(α+{\pi\over 6}\right)+...+sin\left(α+(k-1){\pi\over 6}\right)+sin\left(α+{k\pi\over 6}\right)\)
\(={sin\left(α+{(k-1)\pi\over 12}\right).sin{k\pi\over 12}\over sinm\left(\pi\over 12\right)}=sin\left(α+k{\pi\over 6}\right)\)
\(={1\over sin\left(\pi\over 12\right)}\left[ sin\left(α+(k-1){\pi\over12}\right).sin{k\pi\over 12}+sin\left(α+{k\pi\over 6}\right).sin\left(\pi\over 12\right)\right]\)
\(={1\over sin\left(\pi\over 12\right)}\left[ sin\left(α+{k\pi\over12}-{\pi\over12}\right).sin{k\pi\over 12}+sin\left(α+{k\pi\over 6}\right).sin\left(\pi\over 12\right)\right]\)
\(={1\over sin\left(\pi\over 12\right)}\left[\left[sin\left(α+{k\pi\over12}\right).cos{\pi\over12}+cos\left(α+{k\pi\over12}\right)sin\left(\pi\over 12\right)\right]sin{k\pi\over12}+sin\left(k+{k\pi\over 6}\right)sin{\pi\over 12} \right]\)
∵ (sin (A-B)=sinA cos B - cos Asin B)
\(={1\over sin\left(\pi\over 12\right)}\left[\left[sin\left(α+{k\pi\over12}\right).cos{\pi\over12}sin{k\pi\over12}-cos\left(α+{k\pi\over12}\right)sin{\pi\over 12}.sin{\pi\over12}\right]+sin\left(α+{k\pi\over12}\right)+sin\left({\pi\over 12}\right) \right]\)
\(={1\over sin{\pi\over 12}}\left[ sin\left(α+{k\pi\over 12}\right)cos{\pi\over 12}sin{k\pi\over 12}\right]+{sin\left(\pi\over12\right)\over 2}\left[ 2sin\left(α+{k\pi\over 6}\right)-2cos\left(α+{k\pi\over12}\right)sin{k\pi\over12}\right]\)
[Multiply and divide the second term by 2]
\(={1\over sin{\pi\over 12}}\left[ \left[sin\left(α+{k\pi\over12}\right)cos{\pi\over 12}.sin{k\pi\over12}\right] +{sin{\pi\over12}\over2}\left[ 2sin\left(α+{k\pi\over6}\right)-\left(sin\left( α+{k\pi\over12}+{k\pi\over 12}\right)-sin\left(α+{k\pi\over12}-{k\pi\over 12}\right)\right)\right)\right]\)
∵2cos A sin B = sin (A + B) - sin(A-B)
\(={1\over sin{\pi\over 12}}\left[ sin\left(α+{k\pi\over 12}\right)cos{\pi\over 12}sin{k\pi\over 12}+\right]{sin{\pi\over 12}\over 2}\left[2sin\left(α+{k\pi\over 6}\right)-sin\left(α+{k\pi\over6}\right)+sinα\right]\)
\(={1\over sin{\pi\over12}}\left[ \left[ sin\left(α+{k\pi\over 12}\right)cos{\pi\over 12}sin{k\pi\over 12}\right] +{sin{\pi\over 12}\over 2}\left[ sin\left(α+{k\pi\over6}\right)+sin α\right]\right]\)
\(={1\over sin{\pi\over 12}}\left[ sin\left(α+{k\pi\over12}\right)cos{\pi\over 12}sin{k\pi\over 12}+sin{{\pi\over2}\over 2}\left(2sin\left(α+{k\pi\over6}+α\over 2\right)cos\left(α+{k\pi\over 6}-α\over 2\right)\right)\right]\)
\([∵\ sinC+sinD=2sin\left(C+D\over2\right)cos\left[C-D\over 2\right]]\)
\(={1\over sin{\pi\over 12}}\left[ sin\left(α+{k\pi\over 12}\right)cos{k\pi\over 12}+sin{\pi\over 12}.\left[sin\left(α+{k\pi\over 12}\right).cos\left(k\pi\over12\right)\right]\right]\)
\(={1\over sin{\pi\over 12}}.sin\left(α+{k\pi\over12}\right)\left[cos{\pi\over 12}sin{\pi\over 12}+sin{\pi\over 12}cos{k\pi\over 12}\right]\)
\(={sin\left(α+{k\pi\over12}\right).sin(k+1){\pi\over12}\over sin{\pi\over 12}}\) [∵ sin(A + B) = sin A cos B + cos A sin B]
= RHS
∴ p(k + 1) is true
Hence by the principle of mathematical induction, p(n) is true for all values of n.

Let p( n) be the statement \(1^2+2^2+3^2+L+n^2>{n^3\over 3}\)
Step1: Putting n = 1, we get
\(1>{1^3\over 3}⇒1>{1\over 3}\) Which is true
∴ p(l) is true.
Step 2:
Let us Assume that p(k) is true.
∴ \(1^2+2^2+3^2+...+k^2>{k^3\over 3}\)
Step 3:
To prove that p( k + 1) is true.
ie to P.T.  \(1^2+2^2+3^2+ ... k^2+(k+1)^2>{(k+1)^3\over3}\)
Consider 1² + 2² + 3² + ... k² + (k + 1)²
\(>{K^3\over 3}+(k+1)^2\) [using (1)]
\(> {k^3\over 3}+k^2+2k+1\)
\(>{1\over 3}[k^2 + 3k^2 + 6k + 3]\)
[∴ (a + b)³ = a³+ 3a² b + 3 ab² + b³]
\(>{1\over 3}{k+1}^3\)
∴ p(k + 1) is true
Hence, by the principle of mathematical induction,p(n) is true for all values of m.

Let p(n) be the statement
\({1\over 2.5}+{1\over 5.8}+{1\over 8.11}+...+{1\over {(3n-1)(3n+2)}}={n\over 6n+4}\)
Step 1: Putting n = 1, we get
\({1\over 2.5}={1\over 6(1)+4}⇒{1\over 10}={1\over 10}\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
\(∵\ {1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}={K\over 6K+4}\)
Step. 3: To show that p(K + 1) is true
ie to P.T \({1\over 2.5+}+{1\over 5.8}+...+{1\over{(3K-1)(3K+2)}}={K\over 6+4}\)
ie to P.T \({1\over 2.5}+{1\over 5.8}+...+{1\over (3K-1)(3K+2)}+{1\over (3K+2)(3K+5)}={K+1\over 6K+10}\)
LHS = \({1\over 2.5}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over 5.8}+...+{1\over (3k-1)(3k+2)}+{1\over (3K+2)(3k+5)}\)
\(={k\over 6k+4}+{1\over (3k+2)(3k+5)}\) [using (1)]
\(={k\over 2(3k+2)}+{1\over (3k+2)(3k+5)}\)
\(={1\over 3k+2}\left[{k\over 2}+{1\over 3k+5}\right]\)
\(={1\over 3k+2}\left[k(3k+5)+2\over 2(3k+5)\right]={1\over 3k+2}\left[3k^2+5k+2\over 2(3k+5)\right]\)
\(={1\over 2(3k+5)(3k+5)}(k+1)(3k+2)\)
\(={k+1\over 2(3k+5)}={k+1\over 6k+10}=RHS\)

∴ p (k + 1) is line.
Hence, by the principle mathematical induction p(n) is true for all values of n.

Let p(n) be the statement
\(1.2+2.3+...+n(n+1)={n(n+1)(n+2)\over 3}\)
Step 1:
Putting n = 1 we get
\(1.2={1(1+1)(11+2)\over 3}={2(3)\over3}=2⇒2=2\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
∵ \(1.2+2.3+3.4+...+K(K+1)={K(K+1)(K+2)\over3}\)
Step 3: To prove that p(K+1) is true
ie to P.T. \(1.2+2.3+3.4+...+K(K+1)+(K+1)(K+2)={(K+1)(K+2)\over 3}\)
LHS = 1.2 + 2.3 +...+K(K+1)+(K+1)(K+2)
\(={K(K+1)(K+2)\over 2}+(K+1)(K+2)\)
\(=(K+1)(K+2)\left[{K+\over3}+1\right]\)
\(={(K+)(K+2)(K+3)\over 3}=RHS\)
∵ p(K+1) is true.
Hence, by mathematical induction, p(n) is true for all values of n.

The given data can be converted as a route map diagram as follows:

The possible choices for number of routes commuting from A to place C via place b, without using similar mode of transportation are
( B₁, T'₁) ( B₂, T'₂ ) ( B₁, A₁ ) ( B₂, T'₁ ) ( B₂, T'₂ ) ( B₂, A'₁ ) ( T₁, A'₁ ) ( T₂, B'₁ ) ( T₂, A'₁ ) ( A_1, B'₁ ) ( A_1, T'₁ )  and ( A_2, T'₂ ) 
Required number of routes of commuting from place A to place C via place 'B'
\(=(2 \times 3)+(2 \times 2)+(1 \times 3)\)
= 6+4 + 0 = 7
Required number of routes  = 20 - 7 = 13

The repetition of digits is allowed

Hundreds	tens	unit
5	6	3

The unit place can be filled in 3 ways using the digits 1, 3, or 5 since we need 3 digit odd numeric Hundreds place can be filled in 5 ways excluding 0 and repetition of digits is allowed.
Tens place can be filled in 6 ways .
∴ By fundamental principle of multiplication, required number of 3 = digit odd numbers
= 5 \(\times\) 6 \(\times\) 3 = 30 \(\times\) 3 = 90.

(i) Repetition of digit is allowed

4

6

6

Since we are going to find numbers between 100 and 500 it has 3 = digits
The unit place can be filled in 6 ways using the digits 0, 1, 2, 3, 4, 5
The tens place also can be filled in 6 ways since repetition of digits is allowed.
The hundreds place can be filled in 4 ways using the digits 1, 2, 3, 4 [excluding 0 and 5]
∴ By fundamental principle of multiplication, required number of 3 - digit numbers = 4 \(\times\) 6 \(\times\) 6 = 144.
(ii) Repetition of digits is not allowed.

4

5

4

Hundreds place can be filled in 4 ways excluding 0 and 5
Tens place can be filled in 5 ways since repetition of digits is not allowed
Unit place can be filled in 4 ways.
∴ By fundamental principle of multiplication, required number of three-digit numbers = 4 \(\times\) 5 \(\times\) 4 = 80.

LHS = cos B - cos C
\(cos\left( \frac { \pi }{ 2 } -C \right) -cosC\)
= sin C - cos C
= \(sinC-\left( 1-2{ sin }^{ 2 }\frac { C }{ 2 } \right) \)
= \(2sin\frac { C }{ 2 } cos\frac { C }{ 2 } -1+2{ sin }^{ 2 }\frac { C }{ 2 } \)
= \(-1+2sin\frac { C }{ 2 } \left[ cos\frac { C }{ 2 } +sin\frac { C }{ 2 } \right] \)
= \(-1+2sin\frac { C }{ 2 } \left[ cos\left( 45-\frac { B }{ 2 } \right) +sin\left( 45-\frac { B }{ 2 } \right) \right] \)
= \(-1+2sin\frac { C }{ 2 } \left[ cos45cos\frac { B }{ 2 } +sin45sin\frac { B }{ 2 } +sin45cos\frac { B }{ 2 } -cos45sin\frac { B }{ 2 } \right] \)
= \(-1+2sin\frac { C }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } cos\frac { B }{ 2 } +\frac { 1 }{ \sqrt { 2 } } sin\frac { B }{ 2 } +\frac { 1 }{ \sqrt { 2 } } cos\frac { B }{ 2 } -\frac { 1 }{ \sqrt { 2 } } sin\frac { B }{ 2 } \right] \)
= \(-1+2sin\frac { C }{ 2 } \left[ \frac { 2 }{ \sqrt { 2 } } cos\frac { B }{ 2 } \right] =-1+\frac { 4 }{ \sqrt { 2 } } sin\frac { C }{ 2 } cos\frac { B }{ 2 } \)
= \(-1+\frac { 4\sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } cos\frac { B }{ 2 } sin\frac { C }{ 2 } \)
= \(-1+2\sqrt { 2 } cos\frac { B }{ 2 } sin\frac { C }{ 2 } \) = RHS
Hence proved.

Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

We need to solve the equation N! = 24 \(\times\) N.
For N = 1, 2, 3, 4, N! < 24 \(\times\) N.
For N = 5, we have N! = 5! = 4! \(\times\) 5 = 24N.
For N> 5, we have N! ≥ 5! N > 24 \(\times\) N.
Hence N = 5.