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    Show that:

    (tanx + cot x)(sinx + cosx) = secx + cosecx
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    What have you tried?
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    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
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    so you have

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

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

    Can you combine \frac{sin^2x}{cosx} + cosx to get that
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    (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
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    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)
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    thanks guys, just worked it out
 
 
 
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