To prove
sec^6-tan^6A =
1+3tan^2A*sec^2A.
Put sec^2A = x and tan^2A = y.
Then sec^2A-tan^2A= 1 is also an identity. So x-y = 1. Using these, we
get:
(x^3-y^3) = (x-y)(x^2+xy+y^2) is
andentity.
sec^6A - tan^6A = 1*(sec^4A
+sec^2A*tan^2A+tan^4A) = {(1+tan^2)sec^2A + sec^2A*tan^2A+ }
=
(sec^2A+tan^2Asec^2A +sec^2Atan^2A+(sec^2A-1)tan^2A) =
(3sec^2Atan^2A +sec^A-tan^2A)
= 3tan^2Asec^2A+1, as
sec^A-tan^2A = 1.
Threrefore, sec^6A -tan^6A =
1+3tan^2A*sec^2A.
b)
We use
a^2-b^2 = (a-b) (a+b) and sin^x+ cos^2x = 1
LHS =
(cos^4A-sin^4A = (cos^2A-sin^2A)(cos^2A+sin^2A) ,
=
cos^2A-sin^2A
=cos^2A-(1-cos^2A)
=
cos^2A-1+cos^2A
= 2cos^2A-1 =
RHS.
c)
LHS =
sinA(1+tanA)+cosA(1+cotA)
= sinA+sin^2A/cosA + cosA
+cos^A/sinA
= sinA +(1-cos^2A)/cosA +cosA
+(1-sin^2A)/sinA
= sinA+1/cosA -cosA +cosA +1/sinA -
sinA
= 1/sinA +1/cosA
=
cosecA+secA =
RHS.
=
=
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