JEE Advanced Level Test: Inverse Trigonometry- 3 - NDA MCQ

sin[cot⁻¹(x+1)] = cos(tan⁻¹x)
⇒ sin[sin^−1(1/√1+(x+1)²)] = cos[ cos⁻¹(1/√1+x²)]
⇒ 1/(√1+(x+1)²) = 1/(√1+x²)
⇒ 1/(x²+2x+2) = 1/(√x²+1)
⇒ x²+2x+2 = x²+1
⇒ 2x + 2 = 1
⇒ 2x = 1 − 2
⇒ 2x = −1
⇒ x = −1/2

a² + b² = c²
tan^-1 A + tan^-1 B = tan(A + B)/(1 - AB)
=> tan^-1 (a/(b+c)) + tan^-1 (b/(c+a))
tan^-1 [(a/(b+c) + b/(c+a))/(1 - ab/(b+c)(c+a))]
=> tan^-1(ac + a² + b² + bc)/(bc + ab + c² + ac - ab)
= tan^-1 (ac + bc + a² + b²)/(ac + bc + c²)
= tan^-1 (ac + bc + c²)/(ac + bc + c²)  As we know that {a² + b² = c²}
= tan^-1(1)
= π/4

Given α = cos^-1(⅗)

cos α = 3/5

sin α = ⅘

tan α = 4/3

Also β = tan^-1(⅓)

So tan β = 1/3

tan (α-β) = (tan α – tan β)/(1 + tan α tan β)

= (4/3 – ⅓)/(1 + 4/9)

= 1/(13/9)

= 9/13

So (α-β) = tan^-1 (9/13)

= sin^-1(9/5√10)

Hence option d is the answer.

sin^-1(tanπ/4) - sin^-1(√3/x) - /6 = 0
sin^-1(1) - π/6 = sin^-1(√3/x)
π/2 - π/6 = sin^-1(√3/x)
2π/6 = sin^-1(√3/x)
π/3 = sin^-1(√3/x)
⇒. sin(π/3) = (√3/x)
⇒ (√3/2) = (√3/x)
⇒ (x)^½ = 2
 x = 4

sin^-1 x + cos^-1(1-x) = -sin^-1 x
cos^-1(1-x) = -2sin^-1 x
(1-x) = cos(-2sin^-1 x) ……………………….(1)
Let 2sin^-1 x = t
 x= sin t/2
Cos t = 1 - 2sin² (t/2)
⇒ 1 - 2x^2
From eq(1), we get ⇒ 1- x= 1-2x²
2x² - x = 0
x(2x - 1) = 0
x = 0,½
sin^-1 + cos^-1(1-x) = sin^-1(-x)
0 + 0 + 0 = 0
π/6 + π/3  ⇒ π/2
sin^(-x) = -π/6 not equal to π/2
x = 0, therefore it has 1 solution.

 sin^-1 x implies [-π/2, π/2]
= sin^-1 x + sin^-1 y + sin^-1 z = 3π/2
sin^-1x = π/2      sin^-1y = π/2    sin^-1 z = π/2
x = 1       y = 1    z = 1
(A) = 1+1+1 -(9/(1+1+1))   = 3 - 9/3
⇒ 3 - 3 = 0
(B) = 1+1+1-1-1-1 = 0

α^2 - α -2 > 0
α^2 - 2α + α -2 > 0
α(α-2) +1(α-2) > 0
(α-2) (α+1) > 0
(-∞, -1) U (2,∞)
Therefore α exist for sec^-1α and cosec^-1 α

Therefore a + b = (17 + 6) = 23
a - b = (17-6) = 11
3b = a + 1
=> 3(6) = 18, 
a + 1 => 17 + 1 = 18

cos(−14π/5)=cos.14π/5=cos.4π/5
Hence, cos.1/2cos⁻¹(cos.4π/5)
=cos.2π/5

We know that for the relation

α = tan^-1((2)^½ - 1) = 2tan^-1 tanπ/8
=2 * π/8 
=> π/4 = cos^-1(1/(2)^½)
β = 3(π/4) - π/6 = 7π/12
Therefore, β > α 
⅓ < 1/(2)^½
=> cos^-1(⅓) > cos^-1(/(2)^½) = π/4
So, γ > α 

2x = tan(2tan^-1 a) + 2tan(tan^-1 a + tan^-1 a³)
= 2 tan(tan^-1 a)/[1-{tan(tan^-1 a)}²] + 2tan (tan^-1 (a+a³)/(1-a*a³))
=2a/(1-a²) + 2(a+a³)/(1-a*a³)
= [2a + 2a³ + 2a + 2a³]/(1-a²)(1+a²)
= 4a/(1-a²)(1+a²)
= 4a(1+a²)/(1-a²)(1+a²)
2x = 4a/(1-a²)
2x(1-a²) = 4a
x = 2a/(1-a²)
=> a²x + 2a - x = 0

Let cos⁻¹ x = tan⁻¹ x = y
⇒x=cosy and tany=x
Now,tany=x
siny/cosy = x
⇒(siny)/x=x
⇒siny=x²
siny=cos²y
⇒siny=(1−sin²y)
⇒sin²y+siny⁻¹ = 0
By quadratic formula,
siny = −1±√(1)²−4×1×(−1))/(2 * 1)
​⇒siny=(−1±√5)/2
∵siny∈[−1,1]
∴siny=(√5−1)/2 
⇒sin(cos^−1x) = (√5−1)/2

Σ(n = 1) tan^-1{4n/(n⁴ - 2n + 2)}
=  tan^-1{4n/(1 + (n² - 1)²)}
=  tan^-1{4n/(1 + (n - 1)²(n + 1)²)}
= tan^-1{4n/( (n + 1)² -  (n - 1)²)}
=  Σ(n = 1) tan^-1(n+1)² - tan(n - 1)²
for(n = 1) tan^-1 (4) - tan(0)²
for(n = 2) tan^-1 (9) - tan(1)²
for(n = 3) tan^-1 (16) - tan(2)²
for(n = n-2)  tan^-1(n-1)² - tan(n - 3)²
for(n = n-1) tan^-1(n)² - tan(n - 2)²
for(n = n) tan^-1(n+1)² - tan(n - 1)² we are left with = tan^-1 n + tan¹(n+1) + tan^-1 1
= π/2 + π/2 - π/4
=> 3π/4
=> sec^-1(-(2)^½)