Step I Let P(n) be the given statement.i.e., P(n) :\(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ n-1 }\theta =\frac { { sin2 }^{ n }\theta }{ { 2 }^{ n }sin\theta } \)
Step II For n = 1, we have, LHS = \(cos\theta\) and \(RHS=\frac { sin2\theta }{ { 2 }sin\theta } =\frac { { 2 }sin\theta cos\theta }{ { 2 }sin\theta } =cos\theta \)
\(\therefore\) LHS = RHS  P (1) is true.
Step III Let us assume that P(k) is true. i.e., P(k) : \(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta =\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } \ ...(i)\)
Step IV Now, we shall prove the statement for n = k + 1
For this, we have to show that
\({ cos2 }^{ 2 }\theta ...{ cos2 }^{ (k+1)-1 }\theta \quad \frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta }\)
\(Then \ LHS=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k }\theta \)
\(=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta \ { cos2 }^{ k }\theta\)
\(=\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } .{ cos2 }^{ k }\theta\)
\([Multiplying \ numerator \ and \ denominator \ by \ 2]\)
\(=\frac { { sin2.(2 }^{ k }\theta ) }{ { 2 }^{ k+1 }sin\theta } \ \ [\because 2sin\theta \ cos\theta =sin2\theta ]\)
\(=\frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta } =RHS\)
\(Thus, \ P(k+1) \ is \ true, \ whenever \ P(k) \ is \ true.\)
\(Hence, \ by \ principle \ of \ mathematical \ induction, \ P(n) \ is \ true \ for \ all \ n\in N\)

If elements are not repeated, then number of elements in 
A_{1 \(\cup\)} A_2\(\cup\)   A_3\(\cup\)  ..._\(\cup\)  A₃₀ is 30 \(\times\)5
But each element is used 10 times, so
n(S) = \(\cfrac{30\times5}{10}\) =15
If elements in B₁ ,B₂ ,.....B_n are not r, then total number of elements is 3n but each element is repeated 9 times, so
n(S) = \(\cfrac{3n}{9} \)
\(\Rightarrow 15=\cfrac{3n}{9}\)
n    = 45

We have, g= {(1,1), (2, 3), (3, 5), (4, 7)}
Since, every element has unique image under g. So, g is a function.
Now, g(x)=\(\alpha x+\beta \)
When x = 1, then  g(1)=\(\alpha (1)+\beta \)     ...(i)
\(\Rightarrow 1=\alpha +\beta \)
\(When\quad x=2,\quad then\quad g(2)=\alpha (2)+\beta\)
\(\Rightarrow 3=2\alpha +\beta \)
On solving Eqs.(i) and (ii),we get
\(\alpha =2,\beta =-1\)

\(LHS=(A\cap B')\cup (B\cup C')\)
\(=\{ A'\cup (B')'\} U(B\cup C)\) [By De Morgan's lae]
\(=(A'\cup B)\cup (B\cup C)\)   \([\quad \because (B')'=B]\)
\(=(A'\cup B)\cup B)\cap (A'\cup B)\cup C\)
\(=(A'\cup (B\cup B))\cap (A'\cup (B\cup C))\)
\(=(A'\cup B)\cap (A'\cup B\cup C)\)
\(=(A'\cup B)\quad =RHS\)

Let the two sets be A and B such that n(A) =m and n(B) =n.
Then, number of subsets of set A=2^m
and the number of the subset of set B=2ⁿ
According to given condition, we have
2^m=112+2ⁿ\(\Rightarrow\)2^m-2ⁿ=2⁷-2⁴
On comparing both sides, we get
2^m=2⁷ and 2ⁿ = 2⁴ \(\Rightarrow\)m=7 and n= 4