TRIG Question! Please Help!?

Q2-

cos(x+pi/3)

pi/3 = 180/3 = 60 degree

so,

cos (x+60) = cos x cos 60 - sin x sin 60

= 1/2 cos x - sqrt 3/2 sin x

= 1/2 [ cos x - sqrt 3 sin x]

q 1 Verify that (secx+tanX)(1-sinx)=cosx

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L H S

(sec x + tan x) (1-sin x)

( 1/cos x + sin x/cos x ) ( 1- sin x)

(1+sin x)/cos x (1-sin x)

[ (1+sin x) (1-sin x)] / cos x

(1-sin^2 x)/cos x

cos^2 x/cos x

cos x = R H S

hence verified

(secx+tanX)(1-sinx)=cosx

LHS

= (secx+tanX)(1-sinx)

= (1/cosx+sinx/ cos x)(1-sinx)

= 1/cosx (1+sinx)(1-sinx)

= (1+sinx)(1-sinx)/cos x

= (1-sin^2 x)/cos x

= cos^2 x/cos x

=cosx

*****

cos(x+pi/3)

= cos x cos (pi/3) -sin x sin (pi/3)

= cos x 1/2 -sin x sqrt 3/2

= (1/2 )cos x - (sqrt 3/2) sin x

I can help with the first one...

(secx+tanx)(1-sinx)=cosx

LHS: ((1/cosx)+(sinx/cosx)) * (1-sinx)

=((1+sinx)/cosx) * ((cosx-sinxcosx)/cosx)

=(cosx-sinxcosx+sinxcosx-sin^2cosx)/cos^2x

=(cosx(1-sin^2x))/cos^2x

=cos^3x/cos^2x

=cosx

=RHS

i got here upon trig to be a lot less complicated. Geometry in extreme college develop into the toughest for me, in reality, I nonetheless imagine extreme college geometry is quite harder than truly some the school math education i'm taking to be uncomplicated...I disliked it very a lot.

1. Verify that (secx+tanX)(1-sin x)=cos x

L.H.S = (sec x+tan X)(1-sin x) = (1/cos x + sin x/cos x)(1 - sin x)

= 1/cos x (1+ sin x)(1 - sin x)

= 1/cos x * (1 - sin^2x) = cos^2x/cos x

=cos x = R.H.S (Proved)