Test: Properties Of Inverse Trigonometry - JEE MCQ

3sin^-1(x) = sin^-1(3x - 4x^3) when -1/2<=x<=1/2
Definitely 0.4 comes in this range of x and so

3sin^-1(0.4) = sin^-1[3*0.4 - 4*0.4^3]
3sin^-1(0.4) = sin^-1[1.2 - 4*0.064]
3sin^-1(0.4) = sin^-1[1.2 - 0.256]
3sin^-1(0.4) = sin^-1[0.944]

Finally , sin(3sin^-1(0.4)) = sin{sin^-1(0.944)} = 0.944

Correct Answer :- a

Explanation : cos^-1(12/13) + sin^-1 (3/5)

⇒ sin^-1 5/13 + sin^-1 3/5 

Using the formula,  sin^-1x +  sin^-1y = sin^-1( x√1-y² + y√1-x² )

⇒ sin^-1 ( 5/13√1-9/25 + 3√1-25/169)

⇒ sin^-1 ( 5/13 × 4/5 + 3/5 × 12/13)

⇒ sin^-1 (30 + 36 / 65)

⇒ sin^-1 (56/ 65)

1/(a² - bc)   + 1/(b² - ca)   + 1/(c² - ab)
ab + bc + ca = 0
=> bc = -a(b + c)
=> -bc = a(b+c)
a² - bc = a² + a(b + c) = a(a + b + c)
Similarly b² - ca = b(b + a + c) = b(a + b + c)
c² - ab  = c(a + b + c)
= 1/a(a + b + c)    + 1/b(a + b + c)   + 1/c(a + b + c)
=  bc/abc(a + b + c)  + ac/abc(a + b + c)  + ab/abc(a + b + c)
= (bc + ac + ab)/abc(a + b + c)
= (ab + bc + ac)/abc(a + b + c)
= 0/abc(a + b + c)
= 0

sec^-1x + cosec^-1x =​ π/2 ; x belongs to [ -1 , 1 ]

Because both sin 200⁰ and cos 200⁰ lies in 3rd quadrant. In 3rd quadrant values of both sine and cosine functions are negative.

Let y = tan⁻¹[(a−x)/(a+x)]^1/2
put x = a cos θ
Now, y = tan⁻¹[(a − a cos θ)/(a + a cos θ)]^1/2
⇒ y = tan⁻¹((1 − cos θ)/(1 + cos θ))^1/2
⇒ y = tan⁻¹[[(1−cos θ)(1−cos θ)]/(1+cos θ)(1−cos θ)]^1/2
⇒ y = tan⁻¹[(1−cos θ)^2/1 − cos²θ]^1/2
⇒ y = tan⁻¹[(1 − cos θ)/sin θ]
⇒ y = tan⁻¹[2 sin²(θ/2)/2 sin(θ/2) . cos(θ/2)]
⇒ y = tan⁻¹[tan(θ/2)]
⇒ y = θ/2
⇒ y = 1/2 cos⁻¹(x/a)

(cosx−sinx)/(cosx+sinx) = (1−tanx)/(1+tanx) 
tan(A−B) = (tanA−tanB)/(1 + tanAtanB)
= tan(π/4−x)
putting this value in question.
tan⁻¹tan(π/4−x)
π/4−x.

tan^-1(1-cosx/1+cosx)^½
= tan^-1{(2sin² x/2) / (2cos² x/2)}^½
= tan^-1{(2sin² x/2) / (2cos² x/2)}
= tan^-1(tan x/2)
= x/2

Let, sin^-1(0.6) = A…………(1)
  ( Since, sin² A + cos² A = 1 ⇒ sin² A = 1 - cos² A ⇒ sin A = √(1-cos² A) )

Assume x>0, then
= tan⁻¹ x + tan⁻¹ (1/x)
= tan⁻¹ x + tan⁻¹(1/tan tan⁻¹ x)
= tan⁻¹ x + tan⁻¹ cot tan⁻¹ x
= tan⁻¹ x + tan⁻¹ tan(π/2−tan⁻¹x)
= tan⁻¹ x + π/2 − tan^−1 x   (∵ for x>0, π²−tan⁻¹ x∈(0,π/2))
= π/2.

tan^-1[(3+2x)/(2-3x)]
=> tan^-1[2(3/2 + x)/2(1-3/2x)]
=> tan^-1[(3/2 + x)/(1-3/2x)]
=> tan^-1(3/2) + tan^-1(x)

tan⁻¹ x = θ , so that  x=tanθ . We need to determine  cosθ .
sec²θ = 1 + tan²θ = 1 + x² 
∴s ecθ = ±√(1+x²) 
Then, cos(tan⁻¹x) = cosθ=1/secθ = ±1/√(1+x²)