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Answer:
I have taken A instead of theta..
(2tanAsinA(1-tanA) + 2sinAsec²A)/(1+tanA)²
= 2sinA/(1+tanA)
LHS
= (2tanAsinA(1-tanA) + 2sinAsec²A)/(1+tanA)²
take 2sinA common
LHS
= 2sinA(tanA(1-tanA) + sec²A)/(1+tanA)²
= 2sinA((tanA(1-tanA) + (1+tan²A))/(1+tanA)²
= 2sinA(tanA - tan²A +1 + tan²A)/(1+tanA)²
= 2sinA(1+tanA)/(1+tanA)²
= 2sinA/(1+tanA)
LHS = RHS
hence proved...
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Answer:
Step-by-step explanation:
Consider image..
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