Inverse Trigonometric Functions Test - 6

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^-1 x + sin^-1 y = sin^-1 ( x √1-y ²+ y √1-x ²)

⇒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/(a ²- bc)  + 1/(b ²- ca)  + 1/(c ²- ab)
ab + bc + ca = 0
=>bc = -a(b + c)
=>-bc = a(b+c)
a ²- bc = a ²+ a(b + c) = a(a + b + c)
Similarly b ²- ca = b(b + a + c) = b(a + b + c)
c ²- 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^-1 x + cosec^-1 x =​ 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 −cos² θ]^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⁻¹ tan(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, sin ²A + cos ²A = 1 ⇒sin ²A = 1 - cos ²A ⇒sin A = √(1-cos ²A) )

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⁻¹ x)
= 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=tan θ. We need to determine  cos θ.
sec² θ= 1 + tan² θ= 1 + x²  
∴s ec θ= ±√(1+x² ) 
Then, cos(tan⁻¹ x) = cos θ=1/sec θ= ±1/√(1+x² )