If sin2x=22 - 8√7 Prove that 3(Sinx+Cosx+Tanx+Cosecx+Secx+Cotx) =21
Let sin 2x = k
k = 22- 8√7
k - 22 = -8√7 [Bring 22 to the LHS)
(k-22)² = (-8√7)² [Squaring both sides]
k² - 44k + 484 = 448 [Expanding]
k² - 44k = 448-484 [Bringing 484 to the RHS]
k² - 44k = -36 [Multiply throughout by k]
k³ - 44k² = -36k
k³ - 44k² + 36k = 0 [Bringing -36k to the LHS]
k³ - 49k² +5k² + 8k + 28k = 0 [Rearranging factors]
k³ + 5k² + 8k = 49k² - 28k [Rearranging factors and bring 49k² - 28k to the RHS]
k³ + 4k² + k² + 4k + 4k = 49k² - 28k [Rearranging factors]
k³ + 4k² + k² + 4k + 4k + 4 - 4 = 49k² - 28k [Adding + 4 and -4 to make it a quadratic/polynomial]
k³ + 4k² + k² + 4k + 4k + 4 = 49k² - 28k + 4 [Bringing -4 to the RHS]
k³ + 4k² + k² + 4k + 4k + 4 = (7k - 2)² [Writing 49k² - 28k + 4 in the form of (a-b)²]
(1+k)(k² + 4k + 4) = (7k - 2)² [Factoring LHS]
(1+k)(k + 2)² = (7k - 2)² [Writing k² + 4k + 4 in the form of (a+b)²]
(1+k) = (7k-2)²/(k+2)² [Bringing (k+2)² to the RHS]
(1 + 2sinxcosx) = (7k-2)²/(k+2)² [k = sin2x = 2sinxcosx]
(sin²x + cos²x + 2sinxcosx) = (7k-2)²/(k+2)² [ 1 = sin²x + cos²x and we bring LHS to the form (a+b)²]
(sinx + cosx)² = (7k-2)²/(k+2)²
(sinx + cosx) = (7k-2)/(k+2) [Taking root on both sides]
sinx + cosx = (7k -2)/k / (k+2)/k) [Divide Numerator and Denominator of RHS by k]
sinx + cosx = (7 - 2/k) / (1+2/k)
(sinx + cosx) (1 + 2/k) = 7 - 2/k [Taking (1+2/k) to the LHS]
sinx + cosx + 2sinx/k + 2cosx/k = 7 - 2/k [Multiplying and Expanding]
sinx + cosx + 2/k(sinx + cosx) + 2/k = 7 [Taking Common factor out]
sinx + cosx + 2/k(sinx + cosx + 1) = 7 [Taking Common factor out]
sinx + cosx + 2/k(sinx + cosx + sin²x + cos²x) = 7 [ 1 = sin²x + cos²x]
sinx + cosx + 2/2sinxcos (sinx + cosx + sin²x + cos²x) = 7 [k = sin2x = 2sinxcosx]
sinx + cosx + 1/cosx + 1/sinx + sinx/cosx + cosx/sinx = 7 [Canceling common terms]
sinx+cosx+secx+cosecx+tanx+cotx = 7 [1/cosx=secx,1/sinx=cosecx,sinx/cosx=tan…
Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7
3(sinx + cosx + secx + cosecx + tanx + cotx) = 21
Cheers !!!