Question

solve the equation for the interval [0,2pi)
(secx)^2 - 2 = (tanx)^2

Answer 1

i think there is somethink error with that equation !
please, recheck again ur problem ?

Answer 2

\[\sec ^{2}x-2=\tan ^{2}x\]

Answer 3

yea, i see....
but, do u know that (secx)^2 = (tanx)^2+1
what happend if we convert (secx)^2 to (tanx)^2+1 ?

Answer 4

the problem is # 8 on the attachment I am going to put up

Answer 5

well, from that equation : (secx)^2 - 2 = (tanx)^2
because (secx)^2 = (tanx)^2+1, so left hand side it can be
(tanx)^2+1-2=(tanx)^
(tanx)^2-1=(tanx)^
(tanx)^2-(tanx)^2=1
0=1
so, no solution for x here...

Answer 6

the answer is B

Answer 7

could you look at #6 I got A for it.
Is that correct?

Answer 8

sorry i was answer other member question, wait i will check it

Answer 9

A incorrect

Answer 10

sorry, i was check all option.. no option are correct (if i no mistake calculate it)