simplify trig expression

• Jun 20th 2009, 07:58 AM
Tweety
simplify 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 AM
malaygoel
we have
$\displaystyle secx=\frac{1}{cosx}$

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

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

Quote:

Originally Posted by Tweety

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

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

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

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

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

since $\displaystyle cos^2x+sin^2x=1$
• Jun 20th 2009, 08:05 AM
Tweety
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 AM
skeeter
Quote:

Originally Posted by Tweety
Thanks,

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

could the book be wrong?

you need to take another look at malaygoel's solution.
• Jun 20th 2009, 11:47 AM
Tweety
Quote:

Originally Posted by skeeter
you need to take another look at malaygoel's solution.

ops sorry just realised ,

thanks