Python 3 Solution :

M = 1000000007


def power(x, y):
    x %= M
    ans = 1
    while y > 0:
        if y & 1 == 1:
            ans = ans*x % M
        x = x*x % M
        y >>= 1
    return(ans)


def tan(b, n):
    if n == 1:
        return b
    if n % 2 == 0:
        a = tan(b, n//2)
        return(2*a % M*power((1-a**2 % M+M) % M, M-2) % M)
    else:
        a = tan(b, n-1)
        soorat = (a+b) % M
        makhraj = (1-a*b % M+M) % M
        return(soorat*power(makhraj, M-2) % M)


def solve(p, q, n):
    return(tan(p*power(q, M-2) % M, n))


for _ in range(int(input())):
    p, q, n = map(int, input().split())
    print(solve(p, q, n))

I got the values of final p and q but I didn't get the modulo thing..can anyone help me in this.............

In deriving tan(n * alpha) we can use tangent identity recursively until it reaches tan(alpha).

tan(alpha + beta)
= [tan(alpha) + tan(beta)] / [1 - tan(alpha) * tan(beta)]

The right hand side expanded recursively overflows. To avoid overflow we can ues modular division.

(a / b) mod n
= [(a mod n)(b^(−1) mod n)] mod n

when the right hand side is defined
(that is when b and n are coprime).

To find modular multiplicative inverse of b modulo n, i.e. x below,

bx ≡ 1 mod n

or

bx + ny = gcd(b, n) = 1

use Extended Eucleadean Algorithm that finds quotients x and y in addition to gcd(b, n).

The first and most straightforward approach is to use the addition formula for tangents: This gives us a way to compute given similarly to fast exponentiation: def tan_add(t,u): [// compute tan(a+b) given tan(a) = t and tan(b) = u return (a + b) / (1 - a*b)

def tan(n,t): // compute tan(n*a) given tan(a) = t if n == 1:

        return t
else if n % 2 == 0:
        result = tan(n/2,t)
        return tan_add(result, result)
else:
        result = tan(n-1,t)
     return tan_add(result, t)](http://)

def tan_add(t,u): // compute tan(a+b) given tan(a) = t and tan(b) = u return (a + b) / (1 - a*b)

def tan(n,t): // compute tan(n*a) given tan(a) = t if n == 1: return t else if n % 2 == 0: result = tan(n/2,t) return tan_add(result, result) else: result = tan(n-1,t) return tan_add(result, t)