Test: Inverse Trigonometric Functions (4 August) - 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)
• Similarly, 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)
• as we know, (ab + bc + ac)= 0
• 0/abc(a + b + c)
• 0

Put

sin^-1 x = θ ⇒ x = sin θ

Put,
cos^-1x = θ ⇒ x = cos θ ⇒ cos θ