Tanmay
This page lists questions and answers that were posted by visitors named Tanmay.
Questions
The following questions were asked by visitors named Tanmay.
Answers
The following answers were posted by visitors named Tanmay.
secAcotB-secA-2cotB+2=0 (1/cosA)(cosB/sinB) - (1/cosA)- 2(cosB/sinB) = -2 multiply each term by sinBcosA cosB - sinB - 2cosBcosA = -2cosAsinB cosB - sinB - 2cosBcosA + 2cosAsinB = 0 (cosB - sinB) - 2cosA(cosB - sinB) = 0 (cosB - sinB)(1 - 2cosA) = 0 cosB...
8 years ago
The trick here is to recall that 1 = cos^a+sin^2A cos2A/(1+sin2A) (cos^2A-sin^2A)/(1+2sinAcosA) (cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A) (cosA+sinA)(cosA-sinA)/(cosA+sinA)^2 (cosA-sinA)/(cosA+sinA) (cotA-1)/(cotA+1)
8 years ago
But how?
3 years ago