Given, \(m \sin \theta+n \cos \theta=p\)             .....(i)
\(m \cos \theta-n \sin \theta=q\)                        ......(ii)
On squaring Eq.(i) and Eq.(ii) and adding them we get,
\((m \sin \theta+n \cos \theta)^{2}+(m \cos \theta-n \sin \theta)^{2}=p^{2}+q^{2}\)
Now, simplify the above equation to get the required expression.

\(\frac { \cos { { 45 }^{ ° } } }{ \sec { { 30 }^{ ° } } } +\frac { 1 }{ \sec { { 60 }^{ ° } } } =\frac { \frac { 1 }{ \sqrt { 2 } } }{ \frac { 2 }{ \sqrt { 3 } } } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ \sqrt { 2 } } \times \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } }{ 4 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } +2 }{ 4 } \)

\(3\times 12\times 101+4=4(3\times 3\times 101+1)\)
 = 4(909+1)
 = 4(910)
= a composite number
[∵  Product of more than two factors]

8

Total number of tickets = 40 and multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40
P(a number which is a multiple of 5)
\(=\frac{8}{40}=\frac{1}{5}\)

Here Modal Class = 20-30
i=20,f1=25,,f2=13 , f0=12 , h=10
\(=l+\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\)
\(20+\frac { 25-12 }{ 50-12-13 } \times 10\)
 =20+5.2=25.2

Let \(3+\sqrt { 5 } \) is a rational number
\(3+\sqrt { 5 } =\frac { p }{ q } ,\quad q=0\)
\(3+\sqrt { 5 } =\frac { p }{ q } \)
\(\Rightarrow \sqrt { 5 } =\frac { p }{ q } -3\)
\(\Rightarrow \quad \sqrt { 5 } =\frac { p-3q }{ q } \)
\(\sqrt { 5 } \) is irrational and \(\frac { p-3q }{ q } \) is rational
But rational number cannot be equal to an irrational number.
\(\therefore \quad 3+\sqrt { 5 } \) is an irrational number.

We have tan (3x+30⁰)=1
\(\Rightarrow\) tan (3x + 30⁰)=tan 45⁰   [\(\because\)tan 45⁰ = 1]
\(\Rightarrow\)3x  + 30⁰ = 45⁰
\(\Rightarrow\)3x = 45⁰ - 30⁰ = 15⁰ \(\Rightarrow\) x = 5⁰

We have, \(\frac { \cos { { 60 }^{ 0 } } +\sin { { 45 }^{ 0 } } -\cot { { 30 }^{ 0 } } }{ \tan { { 60 }^{ 0 } } +\sec { { 45 }^{ 0 } } -cosec{ 30 }^{ 0 } } \)
\(=\frac { \frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } -\sqrt { 3 } }{ \sqrt { 3 } +\sqrt { 2 } -2 } =\frac { \frac { \sqrt { 2 } +2-2\sqrt { 2 } \times \sqrt { 3 } }{ 2\sqrt { 2 } } }{ \sqrt { 3 } +\sqrt { 2 } -2 } \)
\([ \because \cos { { 60 }^{ 0 } } =1/2,\sin { { 45 }^{ 0 } } =1/\sqrt { 2 } ,\cot { { 30 }^{ 0 } } =\sqrt { 3 }\)
\( \tan { { 60 }^{ 0 } } =\sqrt { 3 } ,\sec { { 45 }^{ 0 } } =\sqrt { 2 } \quad and\quad cosec{ 30 }^{ 0 }=2 ] \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 2 } (\sqrt { 3 } +\sqrt { 2 } -2) } \)
\(=\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +2\sqrt { 2 } \times \sqrt { 2 } -2\sqrt { 2 } \times 2 } =\frac { \sqrt { 2 } +2-2\sqrt { 6 } }{ 2\sqrt { 6 } +4-4\sqrt { 2 } } \) 

When three coins are tossed simultaneously, all possible outcomes are
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Total number of possible outcomes,n(S)=8
(i) Let E₁ be the event of getting exactly 2 heads.
Then, favourable outcomes, n(E₁)=3
∴ P (getting exactly 2 heads)=P(E₁₎
  \(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 3 }{ 8 } \)
(ii) Let E₂ be the event of getting atmost 2 heads.
Then, E₂=Event of getting 0,1 or 2 heads
So, the favourable outcomes are
{TTT, HTT, THT, TTH, HHT, HTH, THH}.
Number of favourable outcomes, n(E₂)=7
∴ P (getting atmost 2 heads)=P(E₂)
\(=\frac { n(E_{ 2 }) }{ n(S) } =\frac { 7 }{ 8 } \)

Ans. p = 24

x=15, y=20

Given, \(\tan { (A+B) } =\sqrt { 3 } \)
\(\Rightarrow \tan { (A+B) } =\tan { { 60 }^{ 0 } } \)  \([\because \tan { { 60 }^{ 0 } } =\sqrt { 3 } ]\)
\(\therefore \quad A+B={ 60 }^{ 0 }\Rightarrow A={ 60 }^{ 0 }-B\quad \quad \quad ...(i)\)
Also, given \(\tan { (A-B) } =1/\sqrt { 3 } =\tan { { 30 }^{ 0 } } [\because \tan { { 30 }^{ 0 } } =1/\sqrt { 3 } ]\)
\(\Rightarrow A-B={ 30 }^{ 0 }\)      ..(ii)
On substituting the value of B in Eq. (i) in Eq. (ii), we get
\({ 60 }^{ 0 }-B-B={ 30 }^{ 0 }\)
\(\Rightarrow { 60 }^{ 0 }-2B={ 30 }^{ 0 }\Rightarrow { 60 }^{ 0 }-{ 30 }^{ 0 }=2B\\ \Rightarrow \quad { 30 }^{ 0 }=2B\Rightarrow B={ 15 }^{ 0 }\quad \)
On substituting the value of B in Eq. (i), we get
A = 60⁰- 15⁰ = 45⁰
Hence, A = 45⁰ and B = 15⁰   

LHS = \(\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } ++1 }{ (1+\cos { \theta } )\sin { \theta } } \)
\(=\frac { 1+2\cos { \theta } +1 }{ (1+\cos { \theta } )\sin { \theta } } =\frac { 2(\cos { \theta } +1) }{ (1+\cos { \theta } )\sin { \theta } } =2cosec\theta \)