Question

secA-cosA=2,then sec^4A-cos^4=?​

Answer 1

Answer:

33.94 (or) 24√2

Step-by-step explanation:

We know that

SecACosA=1

Squaring on both sides

(SecA -CosA) ^2=(2)^2

Sec^2A+Cos^2A-2SecACosA=4

Sec^2A+Cos^2A-(2)(1)=4

Sec^2A+Cos^2A=4+2

Sec^2A+Cos^2A=6

We know that,

(SecA+CosA) ^2=Sec^2A+Cos^2A+2SecACosA

(SecA+CosA) ^2=6+2

(SecA+CosA) ^2=8

(SecA+CosA) =√8

(SecA+CosA) =2√2

By Formula a^4-b^4=(a^2+b^2)(a^2-b^2)

We can write as (a^2-b^2)= (a+b) (a-b)

By Simplification the Formula can be Written as {(a^4-b^4)=(a^2+b^2)(a+b)(a-b)}

Hear,

(SecA+CosA)=2√2

(Sec^2A+Cos^2A)=6

(SecA-CosA)=2

Sec^4 - Cos^4 =

(Sec^2A+Cos^2A)(SecA-CosA)(SecA+CosA)

Sec^4 - Cos^4 = (6)(2)(2√2)

=(12)(2√2)

Answer = 24√2 (or) 33.94

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