prove that 1-tan²A/1+tan²A= cos²A-sin²A
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Answer:
LHS = 1 - tan²A / 1 + tan²A
= 1- tan²A / sec²A , [ by the formula sec²A = 1+tan²A]
= [1- sin²A/ co²A] / [1/cos²A]
= [cos²A - sin²A] /[cos²A]]/ [1/cos²A]
= [cos²A - sin²A] / cos²A] / [ 1/cos²A]
= [cos²A - sin²A] /cos²A] x [cos²A/ 1]
= [cos²A - sin²A] / 1
= cos²A - sin²A
•°• LHS = RHS
cos²A = sin²A
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