Question Of The Day (4-July-09)

[TestFunda]

Question Of The Day (4-July-09)

[rbh2]

ans is op 3

[rbh2]

use 2/cosx =1/cos(x-y)+1/cos(x+y)
n u will get
cosx secy/2 = +- op 3

[donkeydee]

opt 3

[kirtishekhawat]

not getting exactly guys plz help.....

[manjuc]

one question for u people!

cant we consider the series 1,1,1,1........ as an A.P./H.P/G.P?

for above question if we substitute x=y=0 we get the required answer as 1.

[chinna0105]

Hey why not option 5?????

if i take y=0. they are still in hp....and the question becomes |cosx * 1| = cos x.

if i substitute x=0, it's 1...for x= 60, it's 1/2.....

So, from this the answer turns out to be none of the above....

can any one see any fallacy?

[chinna0105]

Originally Posted by manjuc

one question for u people!

cant we consider the series 1,1,1,1........ as an A.P./H.P/G.P?

for above question if we substitute x=y=0 we get the required answer as 1.

Hey i think, we both are on the same lines... but see my post above.....

[manjuc]

@chinna0105

yes i think ur rite.i overlooked that step.may be its option 5.

but my question is can we consider the series 1,1,1..... as an A.P/H.P/G.P?
and is both arithmetic series and arithmetic progression are same?(i think they r difnt)
and progression means "to progress-increase r decrease"(i thought like that).is it not so?

[implex]

@manjuc!

AP can have a zero common difference
and similalrly for others

the series 1, 1, 1,1 ...
is AP/GP/ HP

So, you have a genuine concern. The fact is either the question setter missed this or he is setting a trap. where in most students will select the option, assuming y to be non zero, which is not to be done.

You cannot assume anything, unless mentioned.

[badman.9001]

the answer is option 3

[nikki23nov]

Originally Posted by rbh2

use 2/cosx =1/cos(x-y)+1/cos(x+y)
n u will get
cosx secy/2 = +- op 3

2/cosx = {cos(x-y) + cos(x+y)} / cos(x-y).cos(x+y)

after this...can u show the solved explanation plz...( m out of touch with those trig. identities..)

[kritikakashyap]

optn 3

[amitmech_bce]

Reciprocal identities





Pythagorean Identities





Quotient Identities





Co-Function Identities





Even-Odd Identities





Sum-Difference Formulas





Double Angle Formulas





Power-Reducing/Half Angle Formulas





Sum-to-Product Formulas





Product-to-Sum Formulas

[ankit4284]

the answer is option 5

[ankit4284]

let me explain everybody nicely yaar

cos(x-y),cos(x),cos(x+y) are in h.p
so there reciprocals are in a.p

hence

1/cos(x-y),1/cos(x),1/cos(x+y) are in a.p

hence

2/cosx=1/cos(x-y) + 1/cos(x+y)

so soving this

2/cosx = cos(x+y) + cos(x-y)/cos(x+y)*cos(x-y)

and cos(x+y).cos(x-y)=(cosx)^2 - (cosy)^2

so the equation we get finaaly is

(cosx)^2(1-cosy)=(cosy)^2

the only values of x and y that satisfies this equation is x=45deg and y=60deg

hence putting these in

mod(cosx * sec(y/2))= sqrt(2/3)

hence the answer is none of these

the answer is option 5

[ankit4284]

or we can also try the substitution of x=y=0 as 1,1,1,1... are in ap and g.p and h.p

so we get option 1 as the answer

may be it is an ambiguous question

[ankit4284]

Originally Posted by ankit4284

let me explain everybody nicely yaar

cos(x-y),cos(x),cos(x+y) are in h.p
so there reciprocals are in a.p

hence

1/cos(x-y),1/cos(x),1/cos(x+y) are in a.p

hence

2/cosx=1/cos(x-y) + 1/cos(x+y)

so soving this

2/cosx = cos(x+y) + cos(x-y)/cos(x+y)*cos(x-y)

and cos(x+y).cos(x-y)=(cosx)^2 - (cosy)^2

so the equation we get finaaly is

(cosx)^2(1-cosy)=(cosy)^2

the only values of x and y that satisfies this equation is x=45deg and y=60deg

hence putting these in

mod(cosx * sec(y/2))= sqrt(2/3)

hence the answer is none of these

the answer is option 5

well i am sorry i missed the identity cos(x-y).cos(x+y)=cosx^2-siny^2
and then the equation is satisfied for x=y=0 because then we may have an ap with the terms 1,1,1,1,1..... and so on with 0 common difference

and hence again the answer may turn out to be option 5 as we can have an ap with 0 common difference

[ankit4284]

and also if we take y to be 0 and x be any value the equation is satisfied and hence

the answer is surely none of these
option 5
not any answer satisfies these constraints

[ankit4284]

but one possible solution to the question may be sqrt(2) as

by solving

(cosx)^2(1-cosy)=siny^2
we can write this as

cosx^2*2sin(y/2)^2=4sin(y/2)^2.cos(y/2)^2

we get

cosx^2=2cos(y/2)^2

cosx^2.sec(y/2)^2=2
and hence
cosx.sec(y/2)=sqrt(2)
hence the answer is coming but this is an ambiguous solution

and hence we cannot consider this as the solution

it is just an another way of manipulating the equation plzzz consider the arguments above and let me know if i am correct