If α is measure of an acute angle and 3sinα = 2cosα, prove that (1-tan²α/1+tan²α)²+(2tanα/1+tan²α)²=1
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We can solve the above question as given in the attachment.
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Given, 3sinα = 2cosα
=> sinα/cosα = 2/3
=> tanα = 2/3 ------(1)
LHS = {(1 - tan²α)/(1 + tan²α)}² + {2tanα/(1 + tan²α)}²
= [{1 - (2/3)²}/{1 + (2/3)²}]²+ [2 × 2/3/{1 + (2/3)²}]²
= {(1 - 4/9)/(1 + 4/9)}² + {4/3/(1 + 4/9)}²
= {(9 - 4)/(9 + 4)}² + {12/(4 + 9)}²
= (5/13)² + (12/13)²
= 25/169 + 144/169
= (25 + 144)/169
= 169/169 = 1 = RHS
=> sinα/cosα = 2/3
=> tanα = 2/3 ------(1)
LHS = {(1 - tan²α)/(1 + tan²α)}² + {2tanα/(1 + tan²α)}²
= [{1 - (2/3)²}/{1 + (2/3)²}]²+ [2 × 2/3/{1 + (2/3)²}]²
= {(1 - 4/9)/(1 + 4/9)}² + {4/3/(1 + 4/9)}²
= {(9 - 4)/(9 + 4)}² + {12/(4 + 9)}²
= (5/13)² + (12/13)²
= 25/169 + 144/169
= (25 + 144)/169
= 169/169 = 1 = RHS
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