How do you prove #(1+cosx)/sinx + sinx/(1+cosx) = 4#?

Question

2852 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-23T14:04:29

#(1+cosx)/sinx+sinx/(1+cosx)=2cscx#

(not #4# as stated in the question)

#(1+cosx)/sinx+sinx/(1+cosx)#

= #((1+cosx)^2+sin^2x)/(sinx(1+cosx)#

= #(1+cos^2x+2cosx+sin^2x)/(sinx(1+cosx)#

= #(1+2cosx+sin^2x+cos^2x)/(sinx(1+cosx)#

=#(2+2cosx)/(sinx(1+cosx)# as #sin^2x+cos^2x=1#

=#(2(1+cosx))/(sinx(1+cosx)#

= #2/sinx# = #2cscx#

Answer 2 · 2016-02-23T15:37:16

The equation has solutions x = 2 arc tan ( 2 #+-# #sqrt#3)

Converting to functions of x/2, the equation reduces to a quadratic in tan x/2. The roots are tan x/2 = 2 #+-# #sqrt#3