The question is ill formed because
Product = 2400
LCM = 96
So HCF = $\frac{product}{LCM} = \frac{2400}{96}=25$
But LCM has to be a multiple of HCF . ths is so because LCM is a multiple of each of the numbers and each number is multiple of HCF
As 25 does not divide 96. so the problem is incorrect, or this case is not possible
as $x^2 + y^ 2 = 1$ we can choose $x = \sin\, t$, $y = \cos\, t$
$3x + 4 y= 3 \sin\, t + 4 \cos\, t$
to convert $3 x + 4y = 3 \sin\,t + 4 \cos\, t$ to the form $A \sin (x+ t)$
$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$
we can choose $3 = 5 \cos\, x$ and $4 = 5 \sin x$ (as $3^2 + 4^2 = 25 = 5^2$
= $5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$
it is maximum when $\sin (x-t) = 1$ and maximum value = 5
minumum when $\sin (x-t) = -1$ and minimum value = 5
working in mod 3 we have $5y^2 = 0 \pmod 3$ or $y =0 \pmod 3$
so y = 3a for some a
similarly x = 5 b for som b
so ge get $75 b^2 + 45 y^2 = 345$
deviding by 15 we get $5b^2 + 3a^2 = 23$
we need to check for $5b^2 < 23$ or $b <=2$
putting b = 1 we get $3a^2 = 18$ or $a^2 = 6$ not an integer
b = 2 gives $3a^3 =3$ or a = 1
so we have a= 1 , b= 2 giving x = 10 and y = 3
We can solve the same in 2 ways. Combinotorics way or alegraic ways
We present here to solve in combinonorics way
From n objects we can choose k objects in $\binom{n}{k}$ ways
let us group the n objects into (n-2,1,1) ways
for choosing k objects this can be done in 3 ways
k objects from n-2 objects that is from 1st set 0 from 2nd set and 0 from3rd set in $\binom{n-2}{k}$ ways
k-1 objects from n-2 objects that is from 1st set 1 from 2nd set or 1 from 3rd set in $2 * \binom{n-2}{k-1}$ ways
k-2 objects from n-2 objects that is from 1st set 1 from 2nd set and 1 from 3rd set in $\binom{n-2}{k-2}$ ways
as all above 3 are mutually exclusive so no of ways =$\binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$
in 2 ways we have computed the number of choosing k objects from n obects so they must be same or
$\binom{n}{k} = \binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$
For it to be group folllowing must be tue.
1) it should be closed
that is if x = a + ib and y = c + id and $(a^2+b^2) = 1$ and $c^2+d^2=1$
and xy = m + ni then $m^2+n^2 =1$
we have $xy = m + ni = (a+ib)(c+id) = (ac - bd) + (bc + ad)i$
we have m = ac - bd and n = bc + ad
$m^2 + n^2 = (ac-bd)^2 + (bc + ad)^2 = a^2c^2 - 2abcbd + b^2d^2 + b^2 c^2 + 2abcd + a^2d^2$
$= a^2c^2 + b^2d^2 + b^2c^2 + a^2d^2 = (a^2+b^2)(c^2 + d^2) = 1$
So it is closed
2) It should have an identity
1 or 1+0i is identity element as $(a+bi)(1+0i) = a+ bi$
3) it should have an inverse
because $a^2+b^2=1$ so it it not zero and hence it has inverse and we need to show that if
m+in is inverse then $m^2+ n^2 =1$ that is the inverse is in this group
$m + in = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a=ib)} = \frac{a-ib}{a^2+b^2} = a - ib$
'so m = a , n = - b and $m^2 + n^2 = a^2 + (-b)^2 = a^2 + b^2 =1$
so it has an inverse
4) assosiativity law holds as unde rcomplex number multiplication assosiativity holds
7p + 4 is a perfect square say $m^2$
so $7p = m^2 -4 = (m+2)(m-2)$
now there are 2 cases
so
2. p is odd
so 7p is odd and $m^2 -4$ is odd
so m+2 and m-2 are co-primes as they differ by 4
so m+2 = 7, m-2 = p gives p = 3 which is prime and m = 5
or m+2 = p and m-2 = 7 giving m = 9 and p =11 which is a prime
so p = 3 or 11
We have
$\cos\, 5t = (\cos\, 5 t + \cos\, t) - \cos\, t$
$= 2 \cos\, 3t \cos\, 2t - \cos\, t$ using $\cos\, A + \cos\, B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$
$= 2 * (4 \cos ^3 t - 3 \cos\, t)(2\cos ^2 t - 1) - \cos\, t$ using formula for $cos 3t$ and $cos 2t$
$= 16 \cos^5 t - 20 \cos^3 t + 6 \cos t -\cos t$
$= 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$
