Question

inverse trigonometry question.....

Answer 1

question no. 21 and 22

Answer 2

@Sachintha

Answer 3

the series in the parentheses are geometric so you can find their sum to infinity using the formula a1 / (1 - r). This will givw us an equation in x to solve.

Answer 4

sin-1[ x / (x - x/2)] + cos-1 [x^2 / (1 - x^2/2)] = pi/2

Answer 5

but the series have alernate + and - sign...
can we use sum of geometric series in such questions?????

Answer 6

OH sorry - theres a mistake there the common ratio in the first one is -1/x not 1/x and also for second r = -1/X^2.

Answer 7

so that should be

sin-1[ x / (x + x/2)] + cos-1 [x^2 / (1 + x^2/2)] = pi/2

Answer 8

oh thanks a lot i know the next prosedure but was stuck in this step

Answer 9

yw - its been a long time since i did these - I'm not sure about the next procedure - so can you tell me! lol!

Answer 10

it is one of the property,
if, sin x + cos y = pi/2
then , x = y

Answer 11

hmmm well i thought i understood tell me more

Answer 12

i think so , there is a correction in the sum of series in arc sin
the sum should be ,
x/( 1+ x/2)
am i correct ?

Answer 13

oh yes that was a typo

Answer 14

next step ..
x / (1 + x/2) = x^2 / ( 1+ X^2/2)
x / ( 2 + x ) = ^2 / ( 2 + x^2 )
after evaluating i got ,
x = 1

Answer 15

can u pls help me in question no. 22

Answer 16

sorry i have to go for 1/2 hour . If you haven't had help by then i will

Answer 17

ok