# Thread: simplify trig expression

1. ## simplify trig expression

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

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

2. we have
$\displaystyle secx=\frac{1}{cosx}$

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

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

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 $\displaystyle cos^2x+sin^2x=1$

3. Thanks,

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

could the book be wrong?

4. 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.

5. Originally Posted by skeeter
you need to take another look at malaygoel's solution.
ops sorry just realised ,

thanks