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.