how do u solve (1-cosX)(1+secX)=tanXsinX?

Answer 1

We have sec(x) = 1/cos(x), tan(x) = sin(x)/cos(x), if cos(x) <>0.

The equation is equivalent to

1 + sec(x) - cos(x) -1 = sec(x) - cos(x) = tan(x) sin(x). Multiplying both sides by cos(x), we get 1 - cos^2(x) = sin^2(x) => sin^2(x) = sin^2( x). So, we have an identity, that is (1-cos(x)(1+sec(x)=tan(x) sin(x) for everyu x such that cos(x) <>0, that is for every x which is not of the form k*pi + pi/2, where k is an integer.

Answer 2

L.H.S.=
(1-cosX)(1+secX)
=1+secx-cosx-cosx.secx
=1+1/cosx-cosx-cosx.1/cosx
=1+1/cosx-cosx-1
=0+1/cosx-cosx
=(1-cos square x)/cosx
(1- cos square x)=sin square x
so, =sin square x / cos x
=(sinx . sinx) / cos x
=sinx/cosx . sinx
sinx/cosx=tanx
so, = tanx.sinx
=R.H.S.

Answer 3

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