what is relation between tan =Sin/cos
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tanA = SinA/Cos A Or L.H.S.
Tan A
=√Sec²A-1 = √1/Cos²A -1 = √1-Cos²A/cos²A
= √sin²A/Cos²A= √tan²A = tanA
Tan A
=√Sec²A-1 = √1/Cos²A -1 = √1-Cos²A/cos²A
= √sin²A/Cos²A= √tan²A = tanA
shizuka26:
thank you but I didn't understand....
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