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) – $$\frac {π}{6}$$