Inverse Trigonometry- 2 - Free MCQ Practice Test with solutions, JEE Chapter

Domain of cos-1x = [-1 ,1] 
Domain of cot-1x = R
Domain of cosec-1x = R - (-1,1)
So taking intersection of domains of all three we have  only {-1 ,1}
As cos-1x is confined between [-1,1] and cosec-1x is not defined between (-1,1), So only -1 and 1 are left.

a) x ∊ [1 , -1]
So cosx has range of -1 to 1 , and cosec-1x is not defined on R - (-1 ,1)
Hence option a is incorrect
 
b) x ∊ R , but cosx has range of -1 to 1 , and cosec-1x is not defined on R - (-1 ,1)
Hence b is incorrect
 
c) x ∊ odd multiple of pi/2
But cosec-1 x is not defined at 0 
 
d) multiple of pie
cosec-1x is defined at 1 and -1
Hence it is correct .

The function f is defined for x = ±1 
Now, f(1) = sin-1(1) + sec-1(1) + tan-1(1) 
= π/2 + 0 + π/4 = 3π/4 
Also, f(-1) = sin-1(-1) + sec-1(-1) + tan-1(-1) 
= - π/2 + π - 0 - π/4 
= 3π/4
Range = { π/4, 3π/4}

f(x) = tanx 
Let  y = tanx
⇒x=tan⁻¹ y
⇒f − 1(x)= tan⁻¹ implies (−1)(1/√(3)
=tan⁻¹(1/√(3)
⇒ =tan⁻¹tan π/6 
=π/6 [∵ tanπ/6 = (1/√(3)]

cos[tan^-1(sin(cot^-1 (3)^½))]
= cos[tan^-1(sin(π/6)]
y = cos[tan^-1(½)] 
Let tan^-1 (½) = θ
tanθ = ½
Applying pythagoras, we get hypotenous = (5)^½
Cosθ = 2/(5)^½
HENCE, y = 2/(5)^½

cos^-1 α1 + cos^-1 α2 + cos^-1 α3 +..........cos^-1αn = 0….(1)
As it is given α(i) = 0
From eq(1), α1 = cos0
α1 = 1, α2 = 1…………………… αn = 1
α1 + α2 + α3 +................+ αn
⇒ 1 + 1 + 1 +..................+ n times
⇒ n

cos^(-1)(x+1)) = cos(tan^-1x)
π/2 - cot^(-1)(x+1) = 2nπ +- tan^-1 x
Put n = 0 
=> π/2 - cot^(-1)(x+1) +- tan^-1 x
= +- tan^-1 x
=> π/2 =  tan^-1 (+-x) + cot^(-1)(x+1)
x+1 = +- x
=> 2x +1 = 0
 x = -½

cot⁻¹ n/π > π/6
n/π < cot(π/6)[as cot⁻¹x is a decreasing function]
n < π×(3)^1/2
​n < π/3
n < 5.16
so max value of n is 5

sin^-1(sin12)+cos^-1(cos12)
Here, we have to convert sin12 into a value from -π/2 to π/2
and cos12 into a value from 0 to π.
We can write,
sin^-1(sin12)+cos^-1(cos12)=sin^-1(sin(12-4π))+cos^-1(cos(4π-12))
=12-4π+4π-12=0
∴sin^-1(sin12)+cos^-1(cos12)=0

x implies(pi/2, pi)
= sin^-1(cos(cos^-1(cosx) + sin^-1(sinx)))
= sin^-1(cos(x + π - x)
= sin^-1(cos π)
= sin^-1(-1)
= -π/2

Correct Answer : d

Explanation : Let α = tan^-1 x

x = tanα

∴ cos2α = 2cos²α−1

= 2(1/(1+x²)^1/2)² - 1

= 2 - (1 + x²)/(1+x²) = (1-x²)/(1+x²)

=> 2tan^-1 x 

=> 2α = cos-1[(1-x²)/(1+x²)]

Let, tan^-1 x = θ         

Therefore, tan θ = x

We know that,

tan 2θ = 2tanθ/(1−tan²θ)

tan 2θ = 2x/(1−x²)

sin 2θ = 2tanθ/(1+tan²θ)

sin 2θ = 2x/(1+x²)

 Let y = tan^-1 x
x = tan y
⇒ 1/x = 1/tan y
⇒ 1/x = cot y
⇒ 1/x = tan(π/2 - y)
⇒ π/2 - y = tan^-1 (1/x)
As we know that y = tan^-1 x
π/2 = tan^-1(x) + tan^-1(1/x)

tan^-1 a + tan^-1 b = tan^-1[(a+b)/(1 - ab)]    (ab>1)
tan^-1[-(a+b)/(1 - ab)] 
π - tan^-1[(a+b)/(1 - ab)]

 f(x)=tan⁻¹.(1−x/1+x),0≤x≤1 ltbr gt 
Now (1−x/1+x) = 2(1+x)−1
Given 0 ≤ x ≤ 1,
⇒2(1+x)−1∈[0,1]
⇒tan⁻¹.(1−x/1+x)∈[tan⁻¹0,tan⁻¹1] 
=> [0,π4]

1/2sin⁻¹[3sin2θ/(5+4cos2θ)]=tan⁻¹x
⇒sin⁻¹[3sin2θ/(5+4cos2θ)]=2tan⁻¹x
⇒sin⁻¹[3(2tanθ/(1+tan²θ)/(5+4(1−tan²θ/1+tan²θ))=2tan⁻¹x
⇒sin⁻¹[6tanθ/(5+5tan²θ+4−4tan²θ]=sin¹(2x/(1+x²))
⇒sin⁻¹[6tanθ/(9+tan2θ)]=sin¹(2x/(1+x²))
⇒sin⁻¹[2(tanθ/3)/1+(tanθ/3)2]=sin¹(2x/(1+x2))
⇒x=1/3(tanθ)

cos⁻¹(cos5)
=2π−5 
as 2π−5 ϵ [0,π]

sin^-1x - cos^-1x = π/6
sin^-1x - cos^-1x = π/2
Adding both the equations, we get
2sin^-1x = 2π/3
sin^-1x = π/3
x = sinπ/3  = [(3)^½]/2
The equation has unique solution.

π22 ≤ sin⁻¹x ≤ π/2 
−π/2 ≤ 2sina ≤ π/2
−π/4 ≤ sina ≤ π4
−1/√2 ≤ a ≤ 1/√2

sin⁻¹ x+sin⁻¹(1−x)=cos⁻¹x
⇒ sin⁻¹x + sin⁻¹
(1−x) = π/2−sin⁻¹x
⇒ sin⁻¹
 (1−x) = π/2−2sin⁻¹x      ...... (i)
Let sin⁻¹ x = y
⇒ x = siny
Therefore, from (i), we get
sin⁻¹(1−x) = π/2 − 2y
⇒1− x = sin(π / 2 − 2y)
⇒ 1 − x = cos2y
⇒ 1 − x = 1 − 2sin²y
⇒ 1−x = 1 − 2x²
⇒ 2x² − x = 0
⇒ x(2x − 1) = 0
⇒ x = 0, 1/2