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 !!!!!!

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)