(sin13°+cos13°)/(sin13°-cos13°)=tanA , what is the value of A?
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sin13°+cos13°/sin13°-cos13° = tan13°+1/tan13°-1
= 1+tan13°/-(1-tan13°)
= -(tan45°+tan13°)/1-(tan45°)(tan13°)
= -(tan(45°+13°))
= -tan58°
= tan(180°-58°)
= tan122°
given rhs = tan A
so tan 122° = tan A
therefore, A = 122°
= 1+tan13°/-(1-tan13°)
= -(tan45°+tan13°)/1-(tan45°)(tan13°)
= -(tan(45°+13°))
= -tan58°
= tan(180°-58°)
= tan122°
given rhs = tan A
so tan 122° = tan A
therefore, A = 122°
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