UnbelievableDreams
  • UnbelievableDreams
Confusing math question
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
ill help you
UnbelievableDreams
  • UnbelievableDreams
Choose the correct expression that completes the identity. sec^2 (x) - tan^2(x)
anonymous
  • anonymous
im sorry this is above my math intelligence

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Nnesha
  • Nnesha
sec^2 = what identity ?
Nnesha
  • Nnesha
do you have sin ,cos identities notes ?
dan815
  • dan815
what are the choices
dan815
  • dan815
|dw:1438637947361:dw|
dan815
  • dan815
|dw:1438637971118:dw|
UnbelievableDreams
  • UnbelievableDreams
No, I don't have notes. I am so confused.
Nnesha
  • Nnesha
now u got the answer if you apply identities that would be easy \[\huge\rm tan^2 +1=\sec^2\]
Nnesha
  • Nnesha
replace sec^2 by tan^2 +1 \[\huge\rm \color{reD}{sec^2(x)}-\tan^2(x)\] \[\huge\rm \color{reD}{tan^2+1 }-\tan^2\]
dan815
  • dan815
u need to know only the basic ones sin^2(x)+cos^2(x)=1 and sec(x)=1/cos(x) csc(x)=1/sin(x) cot(x)=1/tan(x) there are the basic ones to know
UnbelievableDreams
  • UnbelievableDreams
I am still confused. :(
freckles
  • freckles
do you know the Pythagorean identity(ies)?
freckles
  • freckles
\[\sin^2(x)+\cos^2(x)=1 \text{ do you know the one mentioned by dan ?}\]
UnbelievableDreams
  • UnbelievableDreams
No.
freckles
  • freckles
Do you know how to graph the equation x^2+y^2=1? This is the circle with center (0,0) and radius 1. Looks like this: |dw:1438638836144:dw| say we pick a random point (a,b) on the circle: |dw:1438638871637:dw| now we draw a right triangle |dw:1438638917068:dw| we know the relationship between the side of those right triangle to be \[a^2+b^2=1^2 \\ \text{ or } \\a^2+b^2=1 \\ \text{ the 1 comes from the radius of the circle being 1 }\] now we could write this same equation in terms of theta by using the drawing
freckles
  • freckles
\[\cos(\theta)=\frac{a}{1} \text{ or } \cos(\theta)=a \\ \sin(\theta)=\frac{b}{1} \text{ or } \sin(\theta)=b \\ \text{ squaring both sides gives } \\ (\cos(\theta))^2=a^2 \text{ also written as } \cos^2(\theta)=a^2 \\ (\sin(\theta))^2=b^2 \text{ also written as } \sin^2(\theta)=b^2 \\ \text{ so replacing the items \in } a^2+b^2=1 \\ \text{ we have } \cos^2(\theta)+\sin^2(\theta)=1 \]
freckles
  • freckles
you can get the identity that nnesha mentioned from this if you divide both sides by cos^2(theta)
freckles
  • freckles
\[(\cos(\theta))^2+(\sin(\theta))^2=1 \\ \text{ dividing both sides by } (\cos(\theta))^2 \\ \frac{(\cos(\theta))^2+(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1}{(\cos(\theta))^2} \\ \text{ separating fraction on \left hand side } \\ \frac{(\cos(\theta))^2}{(\cos(\theta))^2}+\frac{(\sin(\theta))^2}{(\cos(\theta))^2}=\frac{1^2}{(\cos(\theta))^2} \text{ note: notice I replace } 1 \text{ with } 1^2 \\ \text{ now using law of exponents } \\ (\frac{\cos(\theta)}{\cos(\theta)})^2+(\frac{\sin(\theta)}{\cos(\theta)})^2=(\frac{1}{\cos(\theta)})^2 \\ \text{ now there were ratios mentioned by dan that were equivalent \to some you see here }\] for example you know \[\frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta) \\ \text{ and } \frac{1}{\cos(\theta)}=\sec(\theta)\]
freckles
  • freckles
\[\text{ also you know } \frac{\cos(\theta)}{\cos(\theta)}=1 \text{ on it's domain } \\ \text{ so we have } 1^2+(\tan(\theta))^2=(\sec(\theta))^2 \\\ \text{ which can be written as } \\ 1+\tan^2(\theta)=\sec^2(\theta)\]
freckles
  • freckles
this is the Pythagorean identity you want to apply
UnbelievableDreams
  • UnbelievableDreams
Ah, I see.
UnbelievableDreams
  • UnbelievableDreams
Can you help me another one?
freckles
  • freckles
I can try
UnbelievableDreams
  • UnbelievableDreams
It is similar to this question. But I like to see step by step. 1-tan^2x= ___________
freckles
  • freckles
do you want to write it in terms of sec or use difference of squares
freckles
  • freckles
\[a^2-b^2=(a-b)(a+b) \text{ is difference of squares in factored form }\]
freckles
  • freckles
if you want to write it in terms of sec you can use the identity we just mentioned \[1+\tan^2(\theta)=\sec^2(\theta) \\ \text{ subtract } 1 \text{ on both sides } \\ \tan^2(\theta)=\sec^2(\theta)-1 \\ \text {replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1 \]
UnbelievableDreams
  • UnbelievableDreams
The answer is 1/ sec^2x
UnbelievableDreams
  • UnbelievableDreams
How do you get sec?
freckles
  • freckles
did you replace tan^2(theta) with sec^2(theta)-1 ?
freckles
  • freckles
\[1-\tan^2(\theta) \\ \text{ replace } \tan^2(\theta) \text{ with } \sec^2(\theta)-1 \\ 1-(\sec^2(\theta)-1))\]
freckles
  • freckles
distribute and combine like terms but this shouldn't give you 1/sec^2(x)
UnbelievableDreams
  • UnbelievableDreams
Sec^2x - 1 ?
freckles
  • freckles
nope you distribute \[1-\tan^2(x) \\ \text{ replace } \tan^2(x) \text{ with } \sec^2(x)-1 \\ \text{ which gives } \\ 1-(\sec^2(x)-1) \\ \text{ now distribute } \\1-\sec^2(x)+1 \\ \text{ combine like terms } (1+1)-\sec^2(x)\]
UnbelievableDreams
  • UnbelievableDreams
Ah, I see.

Looking for something else?

Not the answer you are looking for? Search for more explanations.