# help to solve: tan50(thita)tan20(thita)=1 math trigonometry

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Express the formula of tangent function as tan(thita)=sin(thita)/cos(thita).

tan50(thita)tan20(thita)=sin50(thita)sin20(thita)/cos50(thita)cos20(thita)

You should be able to transform the products sin50(thita)sin20(thita) and cos50(thita)cos20(thita) in addition of sines or cosines.

sin50(thita)sin20(thita)=(cos(50-20)-cos(50+20))/2

cos50(thita)cos20(thita)=(cos(50+20)+cos(50-20))/2

tan50(thita)tan20(thita)=(cos(30)-cos(70))/2/(cos(70)+cos(30))/2

(reduced denominators):tan50(thita)tan20(thita)=(cos(30thita)-cos(70thita))/(cos(70thita)+cos(30thita))

Use the result in equation tan50(thita)tan20(thita)=1.

(cos(30thita)-cos(70thita))/(cos(70thita)+cos(30thita))=1=>

=>(cos(30thita)-cos(70thita))=(cos(70thita)+cos(30thita))

cos(30thita)-cos(30thita)=cos(70thita)+cos(70thita)

0=2cos70thita =>cos 70thita=0=>70thita=+(cos 0)^-1+2npi=>

=> 70 thita= +pi/2 + 2npi or 70 thita= -pi/2 + 2npi (divide by 70)

thita=+pi/140 + npi/35 or thita=-pi/140 + npi/35

*Answer: thita=+pi/140 + npi/35 or thita=-pi/140 + npi/35*