anonymous
  • anonymous
sin 17 = a so, cot 253+csc253 = ?
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the choice is a. (a-1)/sqrt(1-a^2) b. (1-a) / sqrt (1-a^2) c. (a-1)/sqrt(a^2-1) d. (1-a) / sqrt (a^2-1) e. (-a-1) / sqrt (1-a^2)
anonymous
  • anonymous
(cos253+1)/sin253=(1+cos270cos17+sin270sin17)/(sin270in17-cos270cos17)=(1-a)/(-a)=1-1/a
anonymous
  • anonymous
the answer is b. (1-a)/sqrt(1-a^2) sin 17=a -> cos 253=cos 73=-a. Furthermore, cot 253 + csc 253 = (cos 253+1)/sin253 hence the answer

Looking for something else?

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

More answers

anonymous
  • anonymous
Notice that 270 - 253 = 17, so sin (17) = -cos (253) cot 253 + csc 253 = [cos(253)/sin(253)] + 1/sin(253) = [cos (253) + 1]/sin(253) but cos(253) = -a so: 1-a/sin(253) from the numerator, we know only b or d could be correct. Looking at the denominator, sin(253), we can use the trigonometric identity: sin^2(x) + cos^2(x) = 1 Thus, sin(253) = sqrt[sin^2(253)] = sqrt[1-cos^2(253)] and -cos(253) is a, so the denominator is sqrt(1-a^2) The end equation is 1-a/sqrt(1-a^2) B is the correct answer

Looking for something else?

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