Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 நேர்மாறு முக்கோணவியல் சார்புகள் Ex 4.4 Textbook Questions and Answers, Notes.

## TN Board 12th Maths Solutions Chapter 4 நேர்மாறு முக்கோணவியல் சார்புகள் Ex 4.4

கேள்வி 1.
முதன்மை மதிப்பு காண்க
(i) sec-1$$\left(\frac{2}{\sqrt{3}}\right)$$
(ii) cot-1$$(\sqrt{3})$$
(iii) cosec-1$$(-\sqrt{2})$$
தீர்வு:

(ii)
cot-1$$(\sqrt{3})$$
cot-1$$(\sqrt{3})$$ = θ என்க
⇒ $$\sqrt{3}$$ = cot θ ⇒ tan θ = $$\frac{1}{\sqrt{3}}$$
⇒ tan θ = tan $$\frac{\pi}{6}$$ ⇒ θ = $$\frac{\pi}{6}$$ [∵ $$\frac{\pi}{6}$$ ∈ $$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$$]
∴ cot-1$$(\sqrt{3})$$ = $$\frac{\pi}{6}$$

(iii) cosec-1$$(-\sqrt{2})$$
cosec-1$$(-\sqrt{2})$$ = θ என்க
⇒ $$\sqrt{2}$$ = cosec θ
⇒ sin θ = $$\frac{-1}{\sqrt{2}}$$
⇒ sin θ = –$$\frac{\pi}{4}$$
⇒ sin θ = sin$$\left(\frac{-\pi}{4}\right)$$ [∵ sin(-θ) = -sin θ]
⇒ θ = $$\frac{-\pi}{4}$$ [∵ $$\frac{\pi}{4}$$ ∈ $$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$$\{0} ]

கேள்வி 2.

(i) tan-1$$(\sqrt{3})$$ – sec-1(-2)
(ii) sin-1(1) + cos-1$$\left(\frac{1}{2}\right)$$ + cot-1(2)
(iii) cot-1(1)+ sin-1$$\left(\frac{-\sqrt{3}}{2}\right)$$ – sec-1$$(-\sqrt{2})$$
தீர்வு:
(i) tan-1$$(\sqrt{3})$$ – sec-1(-2)

(ii) sin-1 + cos-1$$\left(\frac{1}{2}\right)$$ + cot-1(2)

= –$$\frac{\pi}{2}$$ + $$\frac{\pi}{3}$$ + cot-1(2)
= cot-1(2) + $$\frac{-3 \pi+2 \pi}{6}$$ = cot-1(2) – $$\frac{\pi}{6}$$

(iii)

⇒ y = –$$\frac{\pi}{3}$$
sec-1$$(-\sqrt{2})$$ = z என்க
⇒ sec z = –$$\sqrt{2}$$ ⇒ cos z = $$\frac{-1}{\sqrt{2}}$$
⇒ cos z = -cos$$\frac{\pi}{4}$$ ⇒ cos z = cos $$\left(\pi-\frac{\pi}{4}\right)$$
⇒ cos z = cos$$\left(\frac{3 \pi}{4}\right)$$ ⇒ z = $$\frac{3 \pi}{4}$$
∴ cot-1(-1) + sin-1$$\left(\frac{-\sqrt{3}}{2}\right)$$ – sec-1(-$$\sqrt{2}$$)
= $$\frac{\pi}{4}$$ – $$\frac{\pi}{3}$$ – $$\frac{3 \pi}{4}$$