Use basic identities to simplify the expression. one divided by cotangent of theta to the second power. + sec θ cos θ

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Use basic identities to simplify the expression. one divided by cotangent of theta to the second power. + sec θ cos θ

Mathematics
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please help me! I am taking precalc online and don't understand a single thing about basic identities
so its \(\huge \dfrac{1}{\cot^2 \theta + \sec \theta \cos \theta} \)
firstly, use the identity that sec theta = 1/cos theta so, sec theta cos theta = 1 and do you know what is \(\large 1+\cot^2 \theta = ... ?\)

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only the cot^2theta is the denominator
no I don't know, I am COMPLETELY lost in this subject
okk., so anyways you can use sec theta cos theta = 1 \(\large \dfrac{1}{\cot^2 \theta }+1 \)
now 1/ cot theta = tan theta so, \(\large \tan^2 \theta+1 \)
and ther's a standard pythagorean identity that \(\large 1+\tan^2 \theta = \sec^2 \theta \) thats it! :)
okay thank you!
can you walk me through another please?
sure
Simplify the expression. quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity secant of x to the power of two plus cosecant of x to the power of two.
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yes
\(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x }\)
ok, we use the identity that csc x = 1/ sin x sec x = 1/cos x
okay
\(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{\sin^2 x \cos^2 x }{\sin^2 x + \cos^2 x }\)
now we can use the famous pythagorean identity \(\Large \color{red}{\sin^2x + \cos^2 x =1 }\)
what remains is your answer! :)
so the answer is just 1? or am I wrong?
denominator = 1 yes what remains is numerator sin^2 x cos^2 x
you have choices for this too ?
yes I do the choices are: sin^2x cos^2x -1 1
im sorry im so stupid. im usually really good at math but precalc just makes no sense to me
oops, i did a simplification error \(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{1 }{\sin^2 x + \cos^2 x } = 1\)
okay and i have a question. next year I planned on taking ap stat. is that anything like precalc because i hope not. is it easier or harder?
if you practice less, math is hard. if you practice more, math is easy. irrespective of any topic/subject
well are the concepts of ap stat very similar to or very different to those of pre-calculus?
statistics is a different branch of math altogether, so very different.
i just think I'm not good at online class taking because I learn better from an actual person rather than a computer screen. I need the physical classroom environment and a teacher to give me instruction.
what kind of problems does ap stat entail? word problems? graphs?
both.
oh okay thank you soooooo much! is there anything else that i should know?
try to take up easy questions before going to harder ones. if you get stuck anytime, OpenStudy is always Open for you :)

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