Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.6 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.6

Choose the most suitable answer from the given four alternatives:

Question 1.

The value of sin^{-1}(cos x), 0 ≤ x ≤ π is

(a) π – x

(b) x – \(\frac {π}{2}\)

(c) \(\frac {π}{2}\) – x

(d) x – π

Solution:

(c) \(\frac {π}{2}\) – x

Hint:

sin^{-1}(cos x) = sin^{-1}(sin(\(\frac {π}{2}\) – x)) = \(\frac {π}{2}\) – x

Question 2.

If sin^{-1} x + sin^{-1} y = \(\frac {2π}{3}\); then cos^{-1} x + cos^{-1} y is equal to

(a) \(\frac {2π}{3}\)

(b) \(\frac {π}{3}\)

(c) \(\frac {π}{6}\)

(d) π

Solution:

(b) \(\frac {π}{3}\)

Hint:

sin^{-1}x + cos^{-1}x + cos^{-1}y + sin^{-1}y = \(\frac {π}{2}\) + \(\frac {π}{2}\) = π

\(\frac {2π}{3}\) + cos^{-1}x + cos^{-1}y = π

cos^{-1}x + cos^{-1}y = π – \(\frac {2π}{3}\) = \(\frac {3π-2π}{3}\) = \(\frac {π}{3}\)

Question 3.

sin^{-1}\(\frac {3}{5}\) – cos^{-1}\(\frac {12}{13}\) + sec^{-1}\(\frac {5}{3}\) – cosec^{-1}\(\frac {13}{12}\) is equal to

(a) 2π

(b) π

(c) 0

(d) tan^{-1}\(\frac {12}{65}\)

Solution:

(c) 0

Hint:

Question 4.

If sin^{-1} x = 2sin^{-1} α has a solution, then

(a) |α| ≤ \(\frac {1}{√2}\)

(b) |α| ≥ \(\frac {1}{√2}\)

(c) |α| < \(\frac {1}{√2}\)

(d) |α| > \(\frac {1}{√2}\)

Solution:

(a) |α| ≤ \(\frac {1}{√2}\)

Hint:

If sin^{-1} x = 2sin^{-1} α has a solution then

–\(\frac {π}{2}\) ≤ 2sin^{-1}α ≤ \(\frac {π}{2}\)

–\(\frac {π}{4}\) ≤ sin^{-1}α ≤ \(\frac {π}{4}\)

sin(\(\frac {-π}{4}\)) ≤ α ≤ sin\(\frac {π}{4}\)

–\(\frac {1}{√2}\) ≤ α ≤ \(\frac {1}{√2}\)

-|α| ≤ \(\frac {1}{√2}\)

Question 5.

sin^{-1}(cos x) = \(\frac {π}{2}\) – x is valid for

(a) -π ≤ x ≤ 0

(b) 0 ≤ x ≤ π

(c) –\(\frac {π}{2}\) ≤ x ≤ \(\frac {π}{2}\)

(d) –\(\frac {π}{4}\) ≤ x ≤ \(\frac {3π}{4}\)

Solution:

(b) 0 ≤ x ≤ π

Hint:

sin^{-1} (cosx) = \(\frac {π}{2}\) – x is valid for

cos x = sin (\(\frac {π}{2}\) – x)

cos x ∈ [0, π]

∴ 0 ≤ x ≤ π

Question 6.

If sin^{-1} x + sin^{-1} y + sin^{-1} z = \(\frac {3π}{2}\), the value of show that x^{2017} + y^{2018} + z^{2019} – \(\frac {9}{x^{101}+y^{101}+z^{101}}\) is

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

(a) 0

Hint:

Question 7.

If cot^{-1} x = \(\frac {2π}{5}\) for some x ∈ R, the value of tan^{-1} x is

(a) –\(\frac {π}{10}\)

(b) \(\frac {π}{5}\)

(c) \(\frac {π}{10}\)

(d) –\(\frac {π}{5}\)

Solution:

(c) \(\frac {π}{10}\)

Hint:

tan^{-1} x + cos^{-1} \(\frac {π}{2}\)

tan^{-1}x = \(\frac {π}{2}\) – cot^{-1} x = \(\frac {π}{2}\) – \(\frac {2π}{5}\)

= \(\frac {5π-4π}{10}\) = \(\frac {π}{10}\)

Question 8.

The domain of the function defined by f(x) = sin^{-1} \(\sqrt {x-1}\) is

(a) [1, 2]

(b) [-1, 1]

(c) [0, 1]

(d) [-1, 0]

Solution:

(a) [1, 2]

Hint:

f(x) = sin^{-1} \(\sqrt {x-1}\)

\(\sqrt {x-1}\) ≥ 0

-1 ≤ \(\sqrt {x-1}\) ≤ 1

∴ 0 ≤ \(\sqrt {x-1}\) ≤ 1

0 ≤ x – 1 ≤ 1

1 ≤ x ≤ 2

x ∈ [1, 2]

Question 9.

If x = \(\frac {1}{5}\) the value of cos(cos^{-1}x + 2sin^{-1}x) is

(a) –\(\sqrt{\frac {24}{25}}\)

(b) \(\sqrt{\frac {24}{25}}\)

(c) \(\frac {1}{5}\)

(d) –\(\frac {1}{5}\)

Solution:

(d) –\(\frac {1}{5}\)

Hint:

cos[cos^{-1}x + sin^{-1}x + sin^{-1}x] = cos(\(\frac {π}{2}\) + sin^{-1}x)

= -sin(sin^{-1}x)

[∵ cos(90+θ) = -sin θ]

= -x = –\(\frac {1}{5}\)

Question 10.

tan^{-1}(\(\frac {1}{4}\)) + tan^{-1}(\(\frac {2}{9}\)) is equal to

(a) \(\frac {1}{2}\)cos^{-1}(\(\frac {3}{5}\))

(b) \(\frac {1}{2}\)sins^{-1}(\(\frac {3}{5}\))

(c) \(\frac {1}{2}\)tan^{-1}(\(\frac {3}{5}\))

(d) tan^{-1}(\(\frac {1}{2}\))

Solution:

(d) tan^{-1}(\(\frac {1}{2}\))

Hint:

Question 11.

If the function f(x) = sin^{-1}(x² – 3), then x belongs to

(a) [-1, 1]

(b) [√2, 2]

(c) [-2, -√2]∪[√2, 2]

(d) [-2, -√2]

Solution:

(c) [-2, -√2]∪[√2, 2]

Hint:

-1 ≤ x² – 3 ≤ 1

-1 + 3 ≤ x² ≤ 1 + 3

⇒ 2 ≤ x² ≤ 4

±√2 ≤ x ≤ ± 2

[-2, -√2]∪[√2, 2]

Question 12.

If cot^{-1} 2 and cot^{-1} 3 are two angles of a triangle, then the third angle is

(a) \(\frac {π}{4}\)

(b) \(\frac {3π}{4}\)

(c) \(\frac {π}{6}\)

(d) \(\frac {π}{3}\)

Solution:

(b) \(\frac {3π}{4}\)

Hint:

A + B + C = π (triangle)

cot^{-1} 2 + cot^{-1} 3 + C = π

Question 13.

sin^{-1}(tan\(\frac {π}{4}\)) – sin^{-1}(\(\sqrt{\frac {3}{x}}\)) = \(\frac {π}{6}\). Then x is root of the equation

(a) x² – x – 6 = 0

(b) x² – x – 12 = 0

(c) x² + x – 12 = 0

(d) x² + x – 6 = 0

Solution:

(b) x² – x – 12 = 0

Hint:

Question 14.

sin^{-1}(2 cos²x – 1) + cos^{-1}(1 – 2 sin²x) =

(a) \(\frac {π}{2}\)

(b) \(\frac {π}{3}\)

(c) \(\frac {π}{4}\)

(d) \(\frac {π}{6}\)

Solution:

(a) \(\frac {π}{2}\)

Hint:

sin^{-1}(2 cos² x – 1) + cos^{-1}(1 – 2 sin²x)

= sin^{-1} (2 cos² x – 1) + cos^{-1} (1 – sin² x – sin² x)

= sin^{-1}(2 cos² x – 1) + cos^{-1}(cos² x – (1 – cos²x))

= sin^{-1}(2 cos² x – 1) + cos^{-1}(cos² x – 1 + cos²x)

= sin^{-1}(2 cos² x – 1) + cos^{-1}(2 cos² x – 1)

= \(\frac {π}{2}\) [∵ sin^{-1} x + cos^{-1} x = \(\frac {π}{2}\)]

Question 15.

If cot^{-1}(\(\sqrt {sinα}\)) + tan^{-1}(\(\sqrt {sinα}\)) = u, then cos 2u is equal to

(a) tan²α

(b) 0

(c) -1

(d) tan 2α

Solution:

(c) -1

Hint:

cot^{-1} x + tan^{-1} x = \(\frac {π}{2}\)

∴ u = \(\frac {π}{2}\)

cos 2u = cos 2(\(\frac {π}{2}\)) = cos π = -1

Question 16.

If |x| ≤ 1, then 2 tan^{-1} x – sin^{-1}\(\frac {2x}{1+x²}\) is equal to

(a) tan^{-1}x

(b) sin^{-1}x

(c) 0

(d) π

Solution:

(c) 0

Hint:

sin^{-1}\(\frac {2x}{1+x²}\) = 2 tan^{-1}x

∴ 2 tan^{-1} x – 2 tan^{-1} x = 0

Question 17.

The equation tan^{-1} x – cot^{-1} x = tan^{-1}(\(\frac {1}{√3}\)) has

(a) no solution

(b) unique solution

(c) two solutions

(d) infinite number of solutions

Solution:

(b) unique solution

Hint:

tan^{-1} x – cot^{-1} x = tan^{-1}(\(\frac {1}{√3}\)) …….. (1)

tan^{-1} x – cot^{-1} x = \(\frac {π}{2}\) ……… (2)

Add 1 and 2

2 tan^{-1} x = \(\frac {π}{6}\) + \(\frac {π}{2}\) = \(\frac {2π}{3}\)

tan^{-1} x = \(\frac {π}{3}\)

x = √3 which is uniqe solution.

Question 18.

If sin^{-1} x + cot^{-1}(\(\frac {1}{2}\)) = \(\frac {π}{2}\), then x is equal to

(a) \(\frac {1}{2}\)

(b) \(\frac {1}{√5}\)

(c) \(\frac {2}{√5}\)

(d) \(\frac {√3}{2}\)

Solution:

(b) \(\frac {1}{√5}\)

Hint:

Question 19.

If sin^{-1} \(\frac {x}{5}\) + cosec^{-1}\(\frac {5}{4}\) = \(\frac {π}{2}\), then the value of x is

(a) 4

(b) 5

(c) 2

(d) 3

Solution:

(d) 3

Hint:

Question 20.

sin(tan^{-1} x), |x| < 1 is equal to

(a) \(\frac {x}{\sqrt{1-x^2}}\)

(b) \(\frac {1}{\sqrt{1-x^2}}\)

(c) \(\frac {1}{\sqrt{1+x^2}}\)

(d) \(\frac {x}{\sqrt{1+x^2}}\)

Solution:

(d) \(\frac {x}{\sqrt{1+x^2}}\)

Hint:

tan a = x

W.K.T 1 + tan² a = sec² a

1 + x² = sec² a

sec a = \(\sqrt{1+x^2}\)

\(\frac {1}{cosa}\) = \(\sqrt{1+x^2}\)

cos a= \(\frac {1}{\sqrt{1+x^2}}\)

sin a = \(\sqrt{1-cos^2a}\) = \(\sqrt{1-\frac {1}{1+x^2}}\)

\(\sqrt{\frac{1+x^2 -1}{1+x^2}}\) = \(\frac {x}{\sqrt{1+x^2}}\)