How do you solve #secx-tanx=tan40#?

Question

#secx-tanx=tan40=>x=?#

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Answer 1 · 2017-08-29T21:04:50

#secx-tanx=tan40#

#1/(secx-tanx)=1/tan40#

#[(secx)^2-(tanx)^2]/(secx-tanx)=cot40#

#[(secx+tanx)*(secx-tanx)]/(secx-tanx)=cot40#

#secx+tanx=cot40#

After collecting 2 equations,

#secx-tanx+secx+tanx=tan40+cot40#

#2secx=sin40/cos40+cos40/sin40#

#2/cosx=[(sin40)^2+(cos40)^2]/(sin40*cos40)#

#2/cosx=1/(1/2*sin80)#

#2/sin(90-x)=2/sin80#

#sin(90-x)=sin80#

#x=10#

I used #(secx)^2-(tanx)^2=1# identity.