Question #72b1b

1 Answer
Feb 18, 2018

No Solutions(Ø)

Explanation:

(1-tanx)^2=sec^2x-2secx
2secx-sec^2x+(1-tanx)^2=0

We know that tan^2x+1=sec^2x,
Therefore,
2secx-2tanx=0
secx-tanx=0
1/cosx-sinx/cosx=0
1/cosx(1-sinx)=0
therefore 1/cosx=0 or 1-sinx=0

-sinx=-1(Assuming that cosx≠0)
sinx=1, therefore x=2npi+pi/2(But cosx≠0)
The roots 2npi+pi/2 doesn't satisfies cosx≠0 for all n∈ZZ

Therefore thereforeNo Solutions(x∈Ø)