# 4 - Inverse Trigonometry Questions Answers

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?

Joshi sir comment

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)

= (π-2) + (π-4) + (6-2π)

= 0

If cos-1x - cos-1(y/2) = α , then 4x2-4xycosα + y2 is equal to ?

Joshi sir comment

according to the given condition

cos-1x  =  cos-1(y/2)  +  a

so x = cos ( cos-1y/2   +  a)

so x = y/2 cosa  -  sin cos-1y/2 + sina

so x = y/2 cosa  - [1-(y2/4)]1/2 + sina

so x - sina - y/2 cosa = -[1-(y2/4)]1/2

now square both side and solve

If  sin-1a + sin-1b+ sin-1c = π , then the value of { a√(1-a2) + b√(1-b2) + c√(1-c2)}

Joshi sir comment

let sin-1a = A ,    sin-1b = B    and    sin-1c= C

so  { a√(1-a2) + b√(1-b2) + c√(1-c2)}   = sinAcosA + sinBcosB + sinCcosC   = 1/2 (sin2A+sin2B+sin2C)

= 1/2 (4sinAsinBsinC)

= 2abc

If    [cot‾1x] + [cos‾1x] = 0 , then complete set of value of x is ( [ * ] is GIF)?

Joshi sir comment These are the graphs for cot-1x and cos-1x

violet for cos and green for cot

from graph it is clear that cos part with integer function will be 0 for cos1<x≤1

and cot part with integer function will be 0 for cot1<x<∞