I have studied this chapter.but I am unable to solve this.Please explain both of these in detail.It will be very kind of you If you solve both of the questions instead of pretending that SORRY WE COULD NOT HANDLE MORE THAN TWO QUESTION IN ONE ASKING: 1].If sin inverse x -cos inverse x =pi/6 then x is- 1.1/2 2.(root 3)/2 3.-1/2 4.0 2].If cos (x-y),cos x and cos(x+y) are in H.P. then Mod cos x.sec y/2 mod equals 1).1 2).2 3).Root 2 4).None of these IT COSTS RS.10 RUPEES TO ASK A SINGLE QUESTION !!!!!!

Asked by Om Prakash Arya | 22nd Jan, 2014, 12:17: AM

Expert Answer:

1.
 
2.

Given  cos(x-y), cosx, cos(x+y). are in HP

So, 1/cos(x-y), 1/cos x, 1/cos (x+y) are in AP

Therefore,

2/ cos x = 1/ cos(x-y) + 1/cos (x+y)

2/cosx = [cos (x+y) + cos (x-y) ] / [cos(x-y) . cos (x+y)]

2/ cosx = 2 cos(x) cos (y) / cos2x - sin2y

1/ cosx = cos(x) cos (y) / cos2x - sin2y

By cross multiplication,

cos2 x . cosy = cos2x - sin2y

cos2x - cos2x . cosy = sin2y = 1 - cos2y

cos2x [1- cosy] = (1 - cosy)(1+ cosy)

cos2x = 1 + cosy

cos2x = 2 cos2(y/2)

cos2x/ cos2(y/2) = 2

cos2x sec2(y/2) = 2

cosx sec(y/2) = +sqrt(2) or -sqrt(2)

|cosx sec(y/2)| = sqrt(2)

Answered by  | 22nd Jan, 2014, 02:22: PM

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