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(-\(\sqrt{2}\))
= \(\frac{\pi}{4}\) – \(\frac{\pi}{3}\) – \(\frac{3 \pi}{4}\)
