If `((k+3)^2+1)/(k+3)=2`, show that, `((k+3)^6+1)/(k+3)^3=2`

If (tan 3A)/(tan A)=k , show that ( sin 3A)/(sin A)= (2k)/(k-1) and hence or otherwise prove that either k gt 3 or k lt 1/3 .

det [[x ^ (k), x ^ (k + 2), x ^ (k + 3) y ^ (k), y ^ (k + 2), y ^ (k + 3) z ^ (k) , z ^ (k + 2), z ^ (k + 3)]] = (xy) (yz) (zx) {(1) / (x) + (1) / (y) + (1) / ( from)}

-1=(6k-3)/(k+1)

If (k-1),(2k+1),(6k+3) are in GP then k=?

If tan 3A /tan A =k , show that sin 3A/sin A = 2k/k-1 an and k cannot Second Method lie between 1/3 and 3.

If 2^(k)-2^(k-1)=8 then the value of k^3 is:

If the lines (x -1)/(-3) = (x - 2)/(2k) = (z - 3)/(2) and (x - 1)/(3k ) = (y -1)/(1) = (z - 6)/(-5) are perpendicular to each other, then the value of k is ,

[(k (k + 1)) / (2)] ^ (2) + (k + 1) ^ (3) = [((k + 1) (k + 2)) / (2) +1] ^ (2)

If sin theta=(1)/(2) (a+(1)/(a)) and sin 3 theta=(k)/(2)(a^(3)+(1)/(a^(3))) , then k+6 is equal to :