Question
( tan theta=frac{sin alpha+cos alpha}{sin alpha-cos alpha} )
( Rightarrow operatorname{dan} theta=frac{sin alpha}{sin alpha}+alpha+1 )
( tan theta=frac{operatorname{sen} alpha+1}{tan alpha-1} )
( Rightarrow tan (tan alpha-1)=operatorname{san} alpha+1 )
2) tand bend - tan ( theta=tan alpha+1 )
( Rightarrow quad tan A t )
( Rightarrow operatorname{san} alpha(operatorname{san} theta-1)=1+operatorname{sen} theta )
( Rightarrow tan alpha=frac{tan theta+1}{tan theta-1} )
s) ( tan alpha=frac{frac{sin theta}{cos theta}+1}{frac{sin theta}{cos theta}-1} )
( Rightarrow ) ben ( alpha=frac{sin theta+cos theta}{sin theta-cos theta} )

tan - Sina + cosa , show that tana = sin a - cos a sin 0 + cos sin -cos
Solution
