Question

can some one help me solve : csc^2xsecx/sec^2x+csc^2x

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

yes

Answer 3

best idea is to first rewrite in terms of sine and cosine. then we will have a bunch of algebra to do. it is then easiest to do the algebra if we write a = cos(x) and b = sin(x) so as not to be confused. ready?

Answer 4

yup

Answer 5

\[\frac{csc^(x)sec(x)}{sec^2(x)+csc^2(x)}\]

Answer 6

that first csc is squared

Answer 7

ok rewriting in terms of sine and cosine we get
\[\frac{\frac{1}{sin^2(x)cos(x)}}{\frac{1}{cos^(x)}+\frac{1}{sin^2(x)}}\]

Answer 8

satellite i need help with the question i posted earlier please

Answer 9

yeah that was a typo. now replace cosine by a and sine by b so we can do the algebra unimpeded.

Answer 10

k give me a minute

Answer 11

ok

Answer 12

\[\frac{\frac{1}{b^2a}}{\frac{1}{a^2}+\frac{1}{b^2}}\]

Answer 13

okay

Answer 14

denominator is
\[\frac{a^2+b^2}{a^2b^2}\]

Answer 15

i mean that is just the denominator. numerator is still
\[\frac{1}{b^2a}\]

Answer 16

now we return to trig for a second. a = cosine and b = sine and
\[\sin^2(x)+\cos^2(x)=1\]

Answer 17

so our denominator is
\[\frac{a^2+b^2}{a^2b^2}=\frac{1}{a^2b^2}\] because
\[a^2+b^2=1\]

Answer 18

yup i gotcha

Answer 19

so we have in total a compound fraction that looks like

\[\frac{\frac{1}{b^2a}}{\frac{1}{a^2b^2}}\]

Answer 20

yup

Answer 21

invert and multiply to get
\[\frac{1}{b^2a}\times \frac{a^2b^2}{1}\]

Answer 22

cancel to get \[a\]

Answer 23

yup i get it
thank you so much

Answer 24

so answer is ...
\[cos(x)\]!

Answer 25

yup

Answer 26

btw it is much easier to work with a and b when you write this out. otherwise the algebra is hard to manipulate. only one trig step in this whole thing was
\[\sin^2(x)+\cos^2(x)=1\] the rest was algebra

Answer 27

thanks