Question
Download Solution PDFகீழே கொடுக்கப்பட்டுள்ள கோவையில் (?) என்ற இடத்தில் என்ன வர வேண்டும்?
\(\sqrt{\text {sec}^4 \theta - \text {tan}^4 \theta - \text {tan}^2 \theta}\) × \(\sqrt{\text {cosec}^4 \theta - \text {cot}^4 \theta - \text {cot}^2 \theta}\) = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFகொடுக்கப்பட்டவை:
\(\sqrt{\text {sec}^4 θ - \text {tan}^4 θ - \text {tan}^2 θ}\) × \(\sqrt{\text {cosec}^4 θ - \text {cot}^4 θ - \text {cot}^2 θ}\) = ?
பயன்படுத்தபட்ட சூத்திரம்:
(1) a4 - b4 = (a2 - b2)(a2 + b2)
(2) sec2θ - tan2θ = 1
(3) cosec2θ - cot2θ = 1
(4) √a × √b = √(ab)
கணக்கீடு:
மேலே உள்ள சமன்பாட்டை இவ்வாறு எழுதலாம்:
⇒ \(\sqrt{(\text {sec}^4 θ - \text {tan}^4 θ - \text {tan}^2 θ) × (\text {cosec}^4 θ - \text {cot}^4 θ - \text {cot}^2 θ)}\) = ?
சமன்பாட்டின் இருபுறமும் வர்க்கப்படுத்தினால்
⇒ (sec4θ - tan4θ - tan2θ) × (cosec4θ - cot4θ - cot2θ) = ?2
⇒ [(sec4θ - tan4θ) - tan2θ] × [(cosec4θ - cot4θ) - cot2θ] = ?2
⇒ [(sec2θ - tan2θ)(sec2θ + tan2θ) - tan2θ] × [(cosec2θ - cot2θ)(cosec2θ + cot2θ) - cot2θ)] = ?2
⇒ [(1)(sec2θ + tan2θ) - tan2θ] × [(1)(cosec2θ + cot2θ) - cot2θ)] = ?2
⇒ [sec2θ + tan2θ - tan2θ] × [cosec2θ + cot2θ - cot2θ)] = ?2
⇒ [sec2θ] × [cosec2θ] = ?2
சமன்பாட்டின் இருபுறமும் வர்க்கப்படுத்தினால்
⇒ √[sec2θ × cosec2θ] = ?
⇒ ? = secθ × cosecθ
∴ தேவையான பதில் secθ × cosecθ.Additional Information
தேர்வுகளில் அடிக்கடி பயன்படுத்தப்படும் சூத்திரங்கள்.
சூத்திர பட்டியல் I:
(1) sin2θ + cos2θ = 1
(2) sec2θ - tan2θ = 1
(3) cosec2θ - cot2θ = 1
சூத்திர பட்டியல் II:
(1) sin4θ - cos4θ = (sin2θ - cos2θ)(sin2θ + cos2θ) = (sin2θ - cos2θ)
(2) sec4θ - tan4θ = (sec2θ - tan2θ)(sec2θ + tan2θ) = (sec2θ + tan2θ)
(3) cosec4θ - cot4θ = (cosec2θ - cot2θ)(cosec2θ + cot2θ) = (cosec2θ + cot2θ)
Last updated on Jun 13, 2025
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