Question
I just had a test, and one question I tried heaps of times and I still couldn't answer. It was "Prove secèsinè + cosecè cosè = 2cosecè "
Can someone please tell me if proving this is possible at all? (But don't tell me how to solve it)
Can someone please tell me if proving this is possible at all? (But don't tell me how to solve it)
Answers
Damon
sin x/cos x + cos x/sin x =
(sin^2 x + cos^2 x)/(sin x cos x )=
1/(sin x cos x)
I do not see how that comes out to 2 cosec x
(sin^2 x + cos^2 x)/(sin x cos x )=
1/(sin x cos x)
I do not see how that comes out to 2 cosec x
Reiny
Usually my first step in proving any trig identity is taking an arbitrary angle to see if it makes it true, say 45 degrees.
Picking up from Damon's first line, your equation is
tanx + cotx = 2/sinx
for x = 45 degrees
LS = 1+1 = 2
RS = 2/.707 which is not LS
so the equation is not an identity.
Picking up from Damon's first line, your equation is
tanx + cotx = 2/sinx
for x = 45 degrees
LS = 1+1 = 2
RS = 2/.707 which is not LS
so the equation is not an identity.