Trigonometry- Related angles

Question

If 13sinB+5=0

180<B<270

If 25cosC-7=0

C belongs to [3/2 π , 2 π]

Find the value of

cosec(180+B) x tan(180+C) — sec(360+B) x cos(-15/4 π)

All divided by

cos(-B) x sin(360-C) + sin(90-C) x cos240

Answer 1

13sinB + 5 = 0 180<B<270 B in quadrant III

sinB = -(5/13) Note: cosB = -(12/13)

25cosC - 7 = 0 [(3π/2),2π] C in quadrant IV

cosC = (7/25) Note: tanC = (-24/7) sin C = (-24/25)

csc(180 + B) = 1/(cos(180 + B) = 1/(cos180•cosB - sin180•sinB)

= 1/((-1)•(-12/13) - (0)•(-5/13))

= 1/(12/13)

csc(180 + B) = 13/12

tan(180 + C) = (tan180) + tanC)/(1 - tan180•tanC)

= ((0) + (-24/7)) /(1- (0)•tanC)

= (-24/7)/(1 - 0)

tan(180 + C) = (-24/7)

sec(360 + B) = 1/(cos(360 + B) = 1/(cos360•cosB - (sin360)•sinB)

= 1/((1)•(-12/13) - (0)(-5/13)

= 1/(-12/13)

sec(360 + B) = (-12/13)

cos (-15π/4) =(√2)/2

cos(-B) = cos B = (-12/13)

sin(360 - C) = sin(360)•cosC - cos(360)•sinC

= 0•cosC - 1•(-24/25)

sin(360 - C) = (24/25)

sin(90-C) = (sin90)•(cosC) - (cos90)•(sinC)

= 1•(7/25) - 0

sin(90 - C) = (7/25)

cos240 = (-1/2)

cosec(180+B) x tan(180+C) — sec(360+B) x cos(-15/4 π)

(13/12)•(-24/7) - (-12/13)(√2/2) = (-338 + 42√2)/91

All divided by

cos(-B) x sin(360-C) + sin(90-C) x cos240

(-23/13)•(24/25) + (7/25)•(-1/2) = (-288/325) + (-7/50) = (-667/390)

30(-338 + 42√2)/4669.

There's a lot of tedious arithmetic at the end, and I don't have a calculator here.

I feel certain that the identities are sound.

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