Question

cosx=tany and coty=tanz then sinx=?

Answer 1

Answer:

Step-by-step explanation:

Note: The question is not complete... there should be another equation Cotz = tanx.

Cosx = Tany  --- (1)

Tanz = Coty ---- (2)

Cotz = Tanx  ---  (3)

Now do (1) * (2) * (3)

=> Cosx * Tanz * Cotz = Tany * Coty * Tanx

=> Cosx = Tanx

=> Cosx = Sinx/Cosx

=> Cos²x = Sinx

=> 1 - Sin²x = Sinx

=> Sin²x +Sinx - 1 = 0

The above is in form of a quadratic equation ax² + bx + c = 0 and we know that x = [-b ±√(b² - 4ac) ] / 2a

=> Sinx = {- 1  ± √[1 - 4(1)(-1)] } / 2(1)

=> Sinx  = -1 ±√5 / 2