cotA/2-tanA/2 is equal to
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cotA/2-tanA/2
= (cosA/2 ÷ sin A/2)- (sinA/2 ÷ CosA/2)
= (cos²A/2-sin²A/2)/sinA/2*cosA/2
we know that cos²x-sin²x=cos2x
=cos2(A/2)/ sinA/2*cosA/2
Multiplying and dividing by 2
=cosA/ sinA/2*cosA/2 *2/2
=2cosA/2 sinA/2*cosA/2
we know that sin2x=sinx*cosx
so
finally =2cosA/sinA
=2cotA
hence cotA/2-tanA/2=2cotA
= (cosA/2 ÷ sin A/2)- (sinA/2 ÷ CosA/2)
= (cos²A/2-sin²A/2)/sinA/2*cosA/2
we know that cos²x-sin²x=cos2x
=cos2(A/2)/ sinA/2*cosA/2
Multiplying and dividing by 2
=cosA/ sinA/2*cosA/2 *2/2
=2cosA/2 sinA/2*cosA/2
we know that sin2x=sinx*cosx
so
finally =2cosA/sinA
=2cotA
hence cotA/2-tanA/2=2cotA
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