Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5

Question 1.

Find the value, if it exists. If not, give the reason for non-existence.

(i) sin^{-1} (cos π)

(ii) tan^{-1}(sin(\(\frac {-5π}{2}\)))

(iii) sin^{-1} [sin 5]

Solution:

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

(iii) sin^{-1} [sin 5]

–\(\frac {π}{2}\) ≤ sin^{-1} 5 ≤ \(\frac {π}{2}\)

-3\(\frac {π}{2}\) ≤ 5 ≤ 2π

–\(\frac {π}{2}\) ≤ 5 – 2π ≤ 0 ≤ \(\frac {π}{2}\)

sin(5 – 2π) = sin 5

sin^{-1} (sin 5) = 5 – 2π

Question 2.

Find the value of the expression in terms of x, with the help of a reference triangle.

(i) sin (cos^{-1}(1 – x))

(ii) cos (tan^{-1} (3x – 1))

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

Solution:

(i) sin (cos^{-1}(1 – x)) = sin [cos^{-1}(adj/hyp)]

(ii) cos (tan^{-1}(3x – 1)) = cos [opp/adj]

Let θ = tan^{-1}(3x – 1)

tan θ = 3x- 1

1 + tan² θ = 1 +(3x – 1)²

sec² θ = 9x² – 6x + 2

sec θ = \(\sqrt{9x² – 6x + 2}\)

cos θ = \(\frac{1}{\sqrt{9x² – 6x + 2}}\)

⇒ cos (tan^{-1}(3x – 1)) = \(\frac{1}{\sqrt{9x² – 6x + 2}}\)

Question 3.

Find the value of

(i) sin^{-1}(cos(sin^{-1}(\(\frac {√3}{2}\))))

(ii) cot(sin^{-1} \(\frac {3}{5}\) + sin^{-1}\(\frac {4}{5}\))

(iii) tan(sin^{-1} \(\frac {3}{5}\) + cot^{-1}\(\frac {3}{2}\))

Solution:

Question 4.

(i) tan^{-1}\(\frac {2}{11}\) + tan^{-1} \(\frac {7}{24}\) = tan^{-1} \(\frac {1}{2}\)

(i) tan^{-1}\(\frac {3}{5}\) + cos^{-1} \(\frac {12}{13}\) = sin^{-1} \(\frac {16}{65}\)

Solution:

Question 5.

Prove that

tan^{-1} x + tan^{-1} y + tan^{-1} z = tan^{-1} (\(\frac {x+y+z-xyz}{1-xy-yz-zx}\))

Solution:

tan^{-1} x + tan^{-1} y + tan^{-1} z

Question 6.

tan^{-1} x + tan^{-1} y + tan^{-1} z = π, show that x + y + z = xyz

Solution:

tan^{-1} x + tan^{-1} y + tan^{-1} z = π

x + y + z – xyz = 0

x + y + z = xyz

Question 7.

Prove that

tan^{-1} x + tan^{-1} \(\frac {2x}{1-x^2}\) = tan^{-1} \(\frac {3x-x^3}{1-3x^2}\), |x| < \(\frac {1}{√3}\)

Solution:

Question 8.

Simplify

tan^{-1} \(\frac {x}{y}\) – tan^{-1} \(\frac {x-y}{x+y}\)

Solution:

Question 9.

(i) sin^{-1} \(\frac {5}{x}\) + sin^{-1} \(\frac {12}{x}\) = \(\frac {π}{2}\)

(ii) 2 tan^{-1} x = cos^{-1} \(\frac {1-a^2}{1+a^2}\) – cos^{-1} \(\frac {1-b^2}{1+b^2}\), a > 0, b > 0

(iii) 2 tan^{-1} (cos x) = tan^{-1} (2 cosec x)

(iv) cot^{-1} x – cot^{-1} (x + 2) = \(\frac {π}{12}\), x > 0

Solution:

sin² x = sin x cos x

⇒ sin x cos x – sin² x = 0

⇒ sin x(cos x – sin x) = 0

sin x = 0 (or) cos x – sin x = 0

⇒ x = nπ, n ∈ Z, (or) cos x = sin x

tan x = 1 = tan \(\frac {π}{4}\)

⇒ x = nπ + \(\frac {π}{4}\), n ∈ Z

⇒ (x + 1)² = 4 + 2√3

⇒ (x + 1)² = 1 + 3 + 2√3

⇒ (x + 1)² = (1 + √3)²

⇒ x + 1 = 1 + √3

∴ x = √3

Question 10.

Find the number of the solutions of the equations

tan^{-1}(x – 1) + tan^{-1} x + tan^{-1}(x + 1) = tan^{-1}3x

Solution:

tan^{-1}(x – 1) + tan^{-1} x + tan^{-1}(x + 1)

= tan^{-1}(x – 1) + tan^{-1}(x + 1) + tan^{-1}x

4x – x³ = 6x – 9x³

8x³ = 2x

8x³ – 2x = 0

2x(x² – 1) = 0

x = 0, x² = 1

x = ±1

Number of solutions are three (0, 1 -1)