Prove the following trigonometric identities:
cot²A(secA-1)/1+sinA=sec²A(1-sinA/1+secA)
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cot²A ( sec A - 1/ 1 + sin A) - sec²A (sin A - 1 / 1 + secA) = 0 proved
Step-by-step explanation:
To prove: cot²A( secA - 1) / 1+ sinA = sec²A ( 1- sinA / 1+ secA) or
cot²A ( secA-1/ 1+sinA) - sec²A (sinA-1 / 1+secA) = 0
Proof:
LHS = cot²A ( secA-1/ 1+sinA) + sec²A (sinA-1 / 1+secA)
= cot²A (SecA - 1) (1 -SinA) / (1-Sin²A) + Sec²A (SinA - 1) (SecA - 1) /(Sec²A - 1)
= Cot²A (SecA - SinA. SecA -1 + SinA) / Cos²A + Sec²A (SinA. SecA - SecA - SinA + 1)/ Tan²A
= 1/Sin²A (SecA - SinA. SecA -1 + SinA) + 1/Sin²A (SinA. SecA - SecA - SinA + 1)
= 0/Sin²A
= 0
= RHS
RHS = LHS
Hence proved.
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