Math, asked by mcravi762, 11 months ago

Prove that SecA(1-sinA) (secA+tanA)=1

Answers

Answered by Anonymous
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 xMuskx
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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