Using the definations of sec, cosec, cot and tan simplify the following expression;

$\displaystyle sec^2x cos^5x + cotx \ cosecx \ sin^4x $

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- Jun 20th 2009, 07:58 AMTweetysimplify trig expression
Using the definations of sec, cosec, cot and tan simplify the following expression;

$\displaystyle sec^2x cos^5x + cotx \ cosecx \ sin^4x $ - Jun 20th 2009, 08:02 AMmalaygoel
we have

$\displaystyle secx=\frac{1}{cosx}$

$\displaystyle cotx=\frac{cosx}{sinx}$

$\displaystyle cosecx=\frac{1}{sinx}$

$\displaystyle =(\frac{1}{cosx})^2.\ cos^5x + \frac{cosx}{sinx} .\ \frac{1}{sinx}. \ sin^4x $

$\displaystyle =cos^3x+cosx\sin^2x$

$\displaystyle =cosx(cos^2x+sin^2x)$

$\displaystyle =cosx$.............ANSWER

since $\displaystyle cos^2x+sin^2x=1$ - Jun 20th 2009, 08:05 AMTweety
Thanks,

thats also the same answer I got but my book says the correct answer is cosx

could the book be wrong? - Jun 20th 2009, 08:25 AMskeeter
- Jun 20th 2009, 11:47 AMTweety