How do you solve #\frac { 1} { \sin y - 1} - \frac { 1} { \sin y + 1} = - \sec y#?

1 Answer

No solution

Explanation:

Combine the left side using LCD = #(sin y)^2-1# i.e.

#1/(siny-1)-1/(siny+1)=(siny+1-(siny-1))/(sin^2y-1)#

or #2/ [(sin^2y-1)#

or #2/-(1-sin^2y)=-2/cos^2y#

The right side is #-1/cosy#

Cross multiply to get

#-2/cos^2y=-1/cosy#

or #-2cosy=-cos^2y#

#cos^2y-2cosy=0#

or #cosy(cosy-2)=0#

but as #cosy-2!=0#

Hence #cosy=0#

i.e. #y=(npi)/2, (3npi)/2#

or in general #y=(2n+1)pi/2#, where #n# is an integer.

But for all these #siny-1# or #siny+1# is undefined,

Hence, no solution.