aieee

no bro

tanx = sinx / cosx //

sin 2 x + 1/4 - sin x = 1 - sin 2 x ...........

2sin 2 x - sin x - 3/4 = 0 ..........

(1 ± √7)/2 ....................................

sin x = 1-√7/2 ...................

but this means sin x is <0 n p in 0 - pi

and other val/ is >1......

where am i goin rong?????

dividing with cosx on both sides, we get

2tanx+2 = √(1+tan²x).............. (provided x≠Π/2)

=> 4tan²x+4+8tanx = 1+tan²x

3tan²x + 8tanx + 3 = 0

=> tanx = (-8 ± 2√7)/6 = (-4 ± √7)/3

correct me if i am wrong !

yup race...............even i got d same

write sin x = tan x/√1+tan²x & cos x = 1/√1+tan²x
(-4-√7)/3 + 1 = (√1+tan²x)/2
right hand side can't be -ve but solveing L.H.S it is -ve, so (-4 - √7)/3
can't be the answer. it was asked in aieee 2006 and the answer given was (-4 - √7)/3. how is it possible........
correct me if i have made a mistake

Akand tan is -ve in second quad

ya sorry.....................deleted tht post....i was bein dumb

wat is wrong in post #6

@ 8x10 regarding #6

secx = ±√1+tan²x ....... positive when x ε Q1 and negative when x ε Q2.

also tanx is positive in Q1 and negative in Q2. so while substituting sinx, we get positive for both quadrants Q1 and Q2; whereas, for cosx we get positive for Q1 and negative for Q2.

u have neglected this sign convention. (be cautious buddy!!!)

my answer (#4) has both the values.

one more thing to be noted here is that, both the values of tanx are negative.

hence sinx + cosx = 1/2 only when x ε Q2

hope the entire discussion is clear now :)

confirmation please.!