Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.4 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.4

Question 1.

Find the principle value of

(i) sec^{-1} (\(\frac {2}{√3}\))

(ii) cot^{-1} (√3)

(iii) cosec^{-1} (-√2)

Solution:

(i) Let sec^{-1} (\(\frac{2}{\sqrt{3}}\)) = θ

⇒ sec θ = \(\frac{2}{\sqrt{3}}\)

⇒ cos θ = \(\frac{\sqrt{3}}{2}\) = cos \(\frac{\pi}{6}\)

⇒ θ = \(\frac{\pi}{6}\)

(ii) cot^{-1} (√3)

cot^{-1} (√3) = θ

√3 = cot θ

cot \(\frac {π}{6}\) = cot θ

θ = \(\frac {π}{6}\)

(iii) cosec^{-1} (-√2)

cosec^{-1} (-√2) = θ

cosec θ = -√2 = -cosec(\(\frac {π}{4}\))

= cosec (-\(\frac {π}{4}\))

θ = –\(\frac {π}{4}\)

Question 2.

Find the value

(i) tan^{-1} (√3) – sec^{-1}(-2)

(ii) sin^{-1}(-1) + cos^{-1}(\(\frac {1}{2}\)) + cot^{-1}(2)

(iii) cot^{-1}(1) + sin^{-1}(-\(\frac {√3}{2}\)) – sec^{-1}(-√2)

Solution:

x = tan^{-1}(√3)

tan x = √3 = tan \(\frac {π}{3}\)

x = \(\frac {π}{3}\)

y = sec^{-1}(-2)

sec y = -2 = -sec \(\frac {π}{3}\)

sec y = sec(π – \(\frac {π}{3}\))

sec y = sec(2\(\frac {π}{3}\))

y = (2\(\frac {π}{3}\))

tan^{-1}(√3) – sec^{-1}(-2) = \(\frac {π}{3}\) – \(\frac {2π}{3}\)

= \(\frac {π – 2π}{3}\) = –\(\frac {π}{3}\)

(ii) sin^{-1}(-1) + cos^{-1}(\(\frac {1}{2}\)) + cot^{-1}(2)

x = sin^{-1}(1)

sin x = -1 = sin(-\(\frac {π}{2}\))

x = –\(\frac {π}{2}\)

y = cos^{-1}(\(\frac {1}{2}\))

cos y = \(\frac {1}{2}\) = cos \(\frac {π}{3}\)

y = \(\frac {π}{3}\)

z = cot^{-1}(2)

cot z = 2

z = cot^{-1}(2) is constant.

sin^{-1}(-1) + cos^{-1}(\(\frac {1}{3}\)) + cot^{-1}(2)

= –\(\frac {π}{2}\) + \(\frac {π}{3}\) + cot^{-1}(2)

= –\(\frac {3π+2π}{6}\) + cot^{-1}(2)

= cot^{-1}(2) – \(\frac {π}{6}\)