Any help with ANY questions would be great!! :) A. (1 - 2cos^2x) / (sinxcosx) =

Question

Any help with ANY questions would be great!! :) A. (1 - 2cos^2x) / (sinxcosx) = tanx - cotx B. (cos^2y) / (1 + 2siny - 3sin^2y) = (1 + siny) / (1+3siny) C. (cot^2x) / (csc^2x + 4cscx - 5) = (1 + sinx) / (1 +5sinx) D. (sin^3x + cos^3x) / (sinx + cosx) = 1 - sinxcosx E. ((cosx) / (secx - 1)) - ((cosx) / (tan^2x)) = cot^2x Any help with ANY questions would be great!! :) A. (1 - 2cos^2x) / (sinxcosx) = tanx - cotx B. (cos^2y) / (1 + 2siny - 3sin^2y) = (1 + siny) / (1+3siny) C. (cot^2x) / (csc^2x + 4cscx - 5) = (1 + sinx) / (1 +5sinx) D. (sin^3x + cos^3x) / (sinx + cosx) = 1 - sinxcosx E. ((cosx) / (secx - 1)) - ((cosx) / (tan^2x)) = cot^2x

Explanation / Answer

a. split it up to 1/(sinxcosx)-2cos2x/(sinxcosx) =(sinx/cosx) - (cosx/sinx) cancel the c2 with the cos on bottom so its 1/(sinxcosx)- 2[ (cosx)/(sinx)] = s/c -c/s add the cos/sin from the right to to the left: 1/(sinxcosx) -(cosx)/(sinx) = sinx/cosx common denominator: (1-cos2x)/(sinxcosx) = sinx/cosx 1-cos2x =sin2x                sin2x/ cosxsinx =sinx/cosx     the sin2x cancells and ur left with  sinx/cosx = sinx/cosx

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.