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(tanx+cotx)2=tan2x+cot2x+2cotxtanx(tan x + cotx)^2 = tan^2x + cot^2x + 2cotxtanx

cotx=1tanxcotx = \frac{1}{tanx} therefore:

tan2x+cot2x+2tanxtanx tan^2 x + cot^2x + \dfrac{2tanx}{tanx} tanx cancel out, replace the other 2 with the trig functions and the +2 goes as well.
Reply 2
isnt 2tanxcotx simplified to 2?
For 8d. use your answer from part c, that's what the "hence" means.

0.51sec2x+cosec2x dx=[tanxcotx]0.51\displaystyle \int^1_{0.5} sec^2x + cosec^2x \ dx = \left[tanx - cotx \right]^1_{0.5} stick the numbers in, answer!

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