Prove that SecA(1-sinA) (secA+tanA)=1
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Answered by
8
Answer:
LHS = SecA(1-SinA) (SecA+TanA)
= (1/CosA) (1-SinA) (1/CosA + SinA/CosA)
= (1-SinA/CosA) (1+SinA/CosA)
= (1- SinA) (1+SinA)/ Cos²A
= 1-Sin²A/Cos²A [using (a+b)(a-b) =a²-
b²]
=Cos²A/Cos²A [using 1- Sin²A =
Cos²A]
= 1 = RHS
Answered by
3
Answer:
LHS = SecA(1-SinA) (SecA+TanA)
= (1/CosA) (1-SinA) (1/CosA + SinA/CosA)
= (1-SinA/CosA) (1+SinA/CosA)
= (1- SinA) (1+SinA)/ Cos²A
= 1-Sin²A/Cos²A [using (a+b)(a-b) =a²-
b²]
=Cos²A/Cos²A [using 1- Sin²A =
Cos²A]
= 1 = RHS
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