JEE Advanced Level Test: Inverse Trigonometry- 1 - NDA MCQ

=> θ = tan^-1 0
=> tan θ = 0

Since, here we are considering only principle solutions

tan-1(1) = π/4

cos-1(-1/2) = 2π/3

sin-1(-1/2) = -π/6

Sol : π/4 +2π/3 -π/6

: 3π/4

L.H.S. and R.H.S. of the given equation are defined if 

Greatest value sin^-11 = π/2

On solving, we get 

From the given equation




[we take only the +ve sign before the square root since 

tan (X + Y) = 33 --- eqn1
X = tan^–1 3
So,
tan X = 3
eqn1 ⇒ 33 = (tanX + tanY)/(i-tanX*tanY)
33 = ( 3 + tanY )/( 1 – 3*tanY)
33 – 99*tanY = 3 + tanY
30 = 100 * tanY
tanY = 0.3
So, Y = tan^-1(0.3)

⇒ x = 1(not possible)

[x] = K, K ∈ Z ⇒ K < x < K + 1, where [x] represents integer part of x.

sin^-1 [cos(cos^-1(cosx) + sin^-1 (sinx))],   where x€(π/2,π)

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

Correct Answer :- D

Explanation : sin(2cos^-1(cot(2tan^-1x)))=0 

2cos^-1(cot(2tan^-1x))=sin^-10=0

cos^-1(cot(2tan^-1x))=0

cot(2tan^-1x)=cos0

cot(2tan^-1x)=1

cot(tan^-1(2x/(1−x²)))=1

(1−x²)/2x=1

1−x²=2x

x²+2x−1=0

x=[−2±√(4+4)]/2

x=−1±√2, +-1

If xy + yz + zx = 1,
Then, 1 – xy – yz – zx = 0 ----- eqn1 
tan^-1x + tan^-1y = tan^-1((x+y)/(1-xy))
tan^-1x + tan^-1y + tan^-1z = tan-1((x+y)/(1-xy)) + tan^-1z
= tan^-1( ( ((x+y)/(1-xy)) + z ) / (1 – ((x+y)*z)/(1+xy)))
= tan^-1((x+y+z-xyz)/(1-xy-yz-zx))
= tan^-1((x+y+z-xyz)/0)  [From eqn1]
= π/2

Cancelling x both sides (One root x=0)
25 = 3*(25-16x²)^1/2 + 4*(25-9x²)^1/2
25 - 3*(25-16x²)^1/2 = 4*(25-9x²)^1/2
Squaring both sides,
625 + 9*(25-16x²) – 2*25*3*(25-16x²)^1/2 = 16*(25-9x²)
625 + 225 – 144x² – 2*25*3*(25-16x²)1/2 = 400 – 144x²
450 = 2*25*3*(25-16x²)^1/2
3 = (25-16x²)^1/2
Squaring both sides
9 = 25 – 16x²
x² = 1
x = ±1
So, roots of the equation are
x = -1, 0, 1
Sum of the roots is zero.

From function it is clear that
(1) x(x + 1) > 0    ∴ Domain of square root function.
(2) x² + x + 1>0   ∴ Domain of square root function. 
Domain of sin^-1 function. From (2) and (3)

The given equation can be written as 
tan^-1(x-1) + tan^-1 (x +1) = tan^-1 3x - tan^-1 x