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C3 trigonometry

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(tanx + cot x)(sinx + cosx) = secx + cosecx
Reply 1
Avatar for TenOfThem
TenOfThem
18
What have you tried?
Reply 2
Avatar for bssjonny
bssjonny
OP
12
i tried multiplying out the brackets and changing the tan to sin/cos and the cot to cos/sin.
then i got sin^2xsecx + sinx + cos x + cos^2xcosecx
Reply 3
Avatar for TenOfThem
TenOfThem
18
so you have

sin2xcosx+cos2xsinx+sinx+cosx\frac{sin^2x}{cosx} + \frac{cos^2x}{sinx} + sinx + cosx

Looking at the answer you know that you need 1cosx\frac{1}{cosx}

Can you combine sin2xcosx+cosx\frac{sin^2x}{cosx} + cosx to get that
Original post by bssjonny
i tried multiplying out the brackets and changing the tan to sin/cos and the cot to cos/sin.
then i got sin^2xsecx + sinx + cos x + cos^2xcosecx


you're almost there
write sin x = (sin^2 x) / sin x
and cos x = (cos^2 x) / cos x
then you can combine the fractions, and using (cos^2 x) + (sin^2 x) = 1 it simplifies to the other formula
Reply 5
Avatar for Dan12
Dan12
What you need to do is find a common denominator for tan x + cot x

i.e. tanx +cot x = (sinx/cosx) + (cosx/sinx)
Cross multiply
(sinx sinx + cosx cosx)/(sinx cos x)
(sin^2 x + cos^2 x)/(sinx cosx)

finally multiply this by (sinx + cos x)
Reply 6
Avatar for bssjonny
bssjonny
OP
12
thanks guys, just worked it out :biggrin:

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