trig identities: sinx-cosx/(sinx)(cosx)=secx-cscx

[kaileibailei]

i need help with these trig problem the first two i need to simplify and the 3rd one verify

1.) (tan^2x)(cscx)(cosx)

2.) 1/sin^2x - 1/tan^2x

3.)sinx-cosx/(sinx)(cosx)=secx-cscx

thanks so much!

 

[o_O]

Re: trig identities help

For the first two, convert everything into terms of sinx and cosx. You should see things starting to simplify.

For the third one, looking on the left side, split the fraction into two fractions and see what you can simplify/cancel:
\(\displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}\)

See if these hints help and if not, come back with any questions :wink:

 

[kaileibailei]

thanks those really helped got any hints for

1/secx+1 - 1/secx-1

and

cosx(cscx-secx) ** i got cotx for this problem but im not sure if its correct

 

[kaileibailei]

or anyone else have any hints for these 2 problems?

 

[o_O]

For the first one, try combining it into one fraction.

For the second, you almost got it. Remember: \(\displaystyle a(b + c) = ab + ac\) Looks like you forgot about \(\displaystyle cosx \cdot secx\)

 

[Deleted member 4993]

thanks those really helped got any hints for

1/secx+1 - 1/secx-1 <-- What is the question - what do you need to do? - simplify? evaluate? what??

Also group your answer with parentheses. As written, the expression equals to zero.Perhaps it looks like

\(\displaystyle \frac{1}{sec(x) + 1} - \frac{1}{sec(x) - 1}\\)

which could be written as

1/[sec(x)+1] - 1/[sec(x)-1]

and

cosx(cscx-secx) ** i got cotx for this problem but im not sure if its correct

It is incorrect - please show your work.