1.त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Conditional Identities in Trigonometry),प्रतिबन्धित त्रिकोणमितीय सर्वसमिकाएँ (Conditional Trigonometrical Identities):

Conditional Identities in Trigonometry

त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Conditional Identities in Trigonometry) के इस आर्टिकल में प्रतिबन्धित सर्वसमिका A+B+C=180° आदि के द्वारा त्रिकोणमितीय सवालों को हल करके समझने का प्रयास करेंगे।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके।यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए।आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:- Trigonometric Ratios of Multiple Angle

2.त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ के उदाहरण (Conditional Identities in Trigonometry Examples):

यदि A+B+C=180° या हो तो सिद्ध कीजिए: [प्र०सं० 18-29]
Example:18. \frac{\sin A+\sin B-\sin C}{\sin A+\sin B+\sin C}=\tan \frac{A}{2} \tan \frac{B}{2}
Solution: \frac{\sin A+\sin B-\sin C}{\sin A+\sin B+\sin C}=\tan \frac{A}{2} \tan \frac{B}{2} \\ \text { L.H.S. } \frac{\sin A+\sin B-\sin C}{\sin A+\sin B+\sin C}\\ =\frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)-\sin C}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C} \\ =\frac{2 \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)-\sin C}{2 \sin \left( \frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C} \\ =\frac{2 \cos \frac{C}{2} \cos \left(\frac{A-B}{2}\right)-2 \sin \frac{C}{2} \cos \frac{C}{2}}{2 \cos \frac{C}{2} \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}} \\ =\frac{2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right]}{2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\sin \frac{C}{2}\right]} \\ =\frac{\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{\pi}{2} -\frac{A+B}{2}\right)} \\ =\frac{\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)} \\ =\frac{2 \sin \left(\frac{\frac{A-B}{2}+\frac{A+B}{2}}{2}\right) \cdot \sin \left(\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right)}{2 \cos \left(\frac{\frac{A-B}{2}+\frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right)} \\ =\frac{\sin \frac{A}{2} \sin \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}} \\=\tan \frac{A}{2} \tan \frac{B}{2}= R.H.S. 
Example:19. \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Solution: \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ \text { L.H.S. } \frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C} \\ =\frac{2 \sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)+\sin 2 C}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C} \\=\frac{2 \sin (A+B) \cos (A-B)+\sin 2 C}{2 \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C} \\ =\frac{2 \sin (\pi-C) \cos (A-B)+ \sin 2 C}{2 \cos \left(\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+\sin C} \\ =\frac{2 \sin C \cos (A-B)+2 \sin C \cos C}{2 \cos \frac{C}{2} \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}} \\ =\frac{2 \sin C[\cos (A-B)+\cos C]}{2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\sin \frac{C}{2}\right]} \\=\frac{2 \sin \frac{C}{2} \cos \frac{C}{2}\left[ \cos (A-B)+\cos (\pi-\overline{A+B}) \right]}{\cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right]} \\ =\frac{\sin \frac{C}{2}[\cos (A-B)-\cos (A+B)]}{\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)} \\ = \frac{2 \sin \frac{C}{2} \times 2 \sin \left(\frac{A-B+A+B}{2}\right) \sin \left(\frac{A+B-A+B}{2}\right)}{2 \cos \left(\frac{\frac{A-B}{2}+\frac{A+B}{2}}{2}\right) \cos \left(\frac{\frac{A-B}{2}-\frac{A+B}{2}}{2}\right)} \\ =\frac{2 \sin \frac{C}{2} \sin A \sin B}{\cos \frac{A}{2} \cos \frac{B}{2}} \\ =\frac{2 \sin \frac{C}{2} \cdot 2 \sin \frac{A}{2} \cos \frac{A}{2} \cdot 2 \sin \frac{B}{2} \cos \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}} [ \because 2 \sin A=2 \sin \frac{A}{2} \cos \frac{A}{2} सूत्र से]
=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Example:20. \sin (B+2 C)+\sin (C+2 A)+\sin (A+2 B)=4 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{C-A}{2}\right) \sin \left(\frac{A-B}{2}\right)
Solution: \sin (B+2 C)+\sin (C+2 A)+\sin (A+2 B)=4 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{C-A}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ \text {L.H.S. } \sin (B+2 C)+\sin (C+2 A)+\sin (A+2 B) \\ =\sin (\pi-A+C)+\sin (\pi-B+A)+\sin (A+2 B) \quad \left[\because B+C=\pi-A, C+A=\pi-B \right] \\ =\sin (\pi-\overline{A-C})+\sin (\pi+A-B)+\sin (A+2 B) \\ =\sin (A-C)-\sin (A-B)+\sin (A+2 B) \\ =2 \cos \left(\frac{A-C+A-B}{2}\right) \sin \left(\frac{A-C-A+B}{2}\right)+\sin (A+2 B) \\ =2 \cos \left(\frac{2 A-B-C}{2}\right) \sin \left(\frac{B-C}{2}\right)+\sin (\pi-C+B) \quad \left[\because A+B=\pi-C\right] \\ =2 \cos \left(\frac{2 A-\overline{B+C}}{2}\right) \sin \left(\frac{B-C}{2}\right)-\sin (B-C) \\ =2 \cos \left(\frac{2 A- B-C}{2}\right) \sin \left(\frac{B-C}{2}\right)-2 \sin \left(\frac{B-C}{2}\right) \cos \left(\frac{B-C}{2}\right) \\=2 \sin \left(\frac{B-C}{2}\right)\left[\cos \left(\frac{2 A-B-C}{2}\right)-\cos \left(\frac{B-C}{2}\right)\right] \\ =2 \sin \left( \frac{B-C}{2}\right) \cdot 2 \sin \left(\frac{\frac{2 A-B-C}{2}+\frac{B-C}{2}}{2}\right) \sin \left( \frac{\frac{B-C}{2}-\frac{2 A-B-C}{2}}{2}\right) \\ =4 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{A-C}{2}\right) \sin \left(\frac{\frac{B-C-2 A+B+C}{2}}{2}\right) \\ =4 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{A-C}{2}\right) \sin \left(\frac{B-A}{2}\right) \\ =4 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{C-A}{2}\right) \sin \left(\frac{A-B}{2}\right)=  R.H.S. 
Example:21. \sin (B+C-A)+\sin (C+A-B) +\sin (A+B-C)=4 \sin A \sin B \sin C
Solution: \sin (B+C-A)+\sin (C+A-B) +\sin (A+B-C)=4 \sin A \sin B \sin C \\ \text { L.H.S. } \sin (B+C-A)+\sin (C+A-B)+\sin (A+B-C) \\ =\sin (\pi-A-A)+\sin (\pi-B-B)+\sin (\pi-C-C) \\ =\sin (\pi-2 A)+\sin (\pi-2 B)+\sin (\pi-2 C) \\ =\sin 2 A+\sin 2 B+\sin 2 C \\ =\sin \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)+\sin 2 C \\ =2 \sin (A+B) \cos (A-B)+\sin 2 C \\ =2 \sin (\pi-C) \cos (A-B)+\sin 2 C \\ =2 \sin C \cos (A-B)+2 \sin C \cos C \\ =2 \sin C[\cos (A-B)+\cos C] \\ =2 \sin C \left[\cos (A-B)+\cos (\pi-\overline{A+B}) \right] \\ =2 \sin C \left[ \cos (A-B)-\cos (A+B) \right] \\ =2 \sin C\left[2 \sin \left(\frac{A-B+A+B}{2}\right) \sin \left(\frac{A+B-A+B}{2}\right)\right] \\ =4 \sin C \sin A \sin B \\ =4 \sin A \sin B \sin C=R.H.S
Example:22. \sin ^2 A+\sin ^2 B-\sin ^2 C=2 \sin A \sin B \sin C
Solution:\sin ^2 A+\sin ^2 B-\sin ^2 C=2 \sin A \sin B \sin C \\ \text{L.H.S.} \sin ^2 A+\sin ^2 B-\sin ^2 C \\ =\sin ^2 A+\sin (B+C) \sin (B-C)
[ \sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B सूत्र से]
=\sin^2 A+\sin (\pi-A) \sin (B-C) \\ =\sin ^2 A+\sin A \sin (B-C) \\ =\sin A \left[\sin A+\sin (B-C)\right] \\ =\sin A\left[\sin (\pi-\overline{B+C})+\sin (B-C) \right] \\ =\sin A\left[\sin (B+C)+\sin (B-C)\right] \\ =\sin A \left[2 \sin \left(\frac{B+C+B-C}{2}\right) \cos \left(\frac{B+C-B+C}{2}\right)\right] \\=2 \sin A \sin B \cos C
Example:23. \cos ^2 A+\cos ^2 B+\cos ^2 C=1-2 \cos A \cos B \cos C
Solution: \cos ^2 A+\cos ^2 B+\cos ^2 C=1-2 \cos A \cos B \cos C \\ \text { L.H.S. } \cos ^2 A+\cos ^2 B+\cos ^2 C \\ =1-\sin ^2 A+\cos ^2 B+\cos ^2 C \\ =1+\cos ^2 B-\sin ^2 A+\cos ^2 C \\ =1+\cos (A+B) \cos (A-B)+\cos ^2 C [ \because \cos ^2 B-\sin ^2 A=\cos ^2(A+B) \cos (A-B) सूत्र से]
=1+\cos (\pi-C) \cos (A-B)+\cos ^2 C \\ =1-\cos C \cos (A-B)+\cos ^2 C \\ =1-\cos C[\cos (A-B)-\cos C] \\ =1-\cos C \left[\cos (A-B)-\cos (\pi-\overline{A+B})\right] \\ =1-\cos C[\cos (A-B)+\cos (A+B)] \\ =1-\cos C \cdot 2 \cos \left(\frac{A-B+A+B}{2}\right) \cos \left(\frac{A-B-A-B}{2}\right) \\ =1-2 \cos C \cos A \cos B \\ =1-2 \cos A \cos B \cos C=R.H.S. 
Example:24. \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2} =1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Solution: \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2} =1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ \text { L.H.S. } \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2} \\ =1-\cos ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2} \\ =1-\left(\cos ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}\right) +\sin ^2 \frac{C}{2} \\ =1-\cos \left(\frac{A}{2}+\frac{B}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right)+\sin ^2 \frac{C}{2}
[ \because \cos (A+B) \cos (A-B)=\cos ^2 A-\sin ^2 B सूत्र से]
=1-\cos \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right)+\sin ^2 \frac{C}{2} \\ =1-\sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)+\sin ^2 \frac{C}{2} \\ =1-\sin \frac{C}{2}\left[\cos \left( \frac{A-B}{2}\right)-\sin \frac{C}{2}\right] \\ =1-\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right] \\ =1-\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right] \\ =1-\sin \frac{C}{2} \cdot 2 \sin \left(\frac{\frac{A-B}{2}+\frac{A-B}{2}}{2}\right) \sin \left(\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right) \\ =1-2 \sin \frac{C}{2} \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \\ =1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}= R.H.S. 

Conditional Identities in Trigonometry

Example:25. \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} =2+2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)
Solution: \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} =2+2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right) \\ \text { L.H.S. } \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} \\=1-\sin ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2} \\ =1+\left(\cos ^2 \frac{B}{2}-\sin ^2 \frac{A}{2}\right)+\cos ^2 \frac{C}{2} \\ =1+\cos ^2\left(\frac{B}{2}+\frac{A}{2}\right) \cos \left(\frac{A}{2}-\frac{B}{2}\right)+\cos ^2 \frac{C}{2}
[ \because \cos (A+B) \cos (A-B)=\cos ^2 B-\sin ^2 A सूत्र से]
=1+\cos \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+\cos ^2 \frac{C}{2} \\ =1+\sin \left(\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-\sin ^2 \frac{C}{2} \\ =2+\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right] \\ =2+\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right] \\ =2+\sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right] \\ =2+\sin \frac{C}{2} \cdot 2 \sin \left(\frac{\frac{A-B}{2}+\frac{A+B}{2}}{2}\right) \sin \left(\frac{\frac{A+B}{2}-\frac{A-B}{2}}{2}\right) \\ =2+2 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2} \\ =2+2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
Example:26. \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2} =1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}
Solution: \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2} =1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \\ \text { L.H.S. } \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2} \\ =\sin ^2 \frac{A}{2}+\sin \left(\frac{B}{2}+\frac{C}{2}\right) \sin \left(\frac{B}{2}-\frac{C}{2}\right) [ \because \sin (A+B) \sin (A-B)=\sin ^2 A-\sin ^2 B सूत्र से]
=\sin ^2 \frac{A}{2}+\sin \left(\frac{\pi}{2}-\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right) \\ =\sin ^2\left(\frac{A}{2}\right)+\cos \left(\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right) \\ =1-\cos ^2 \frac{A}{2}+\cos \left(\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right) \\ =1-\cos \left(\frac{A}{2}\right)\left[\cos \frac{A}{2}-\sin \left(\frac{B-C}{2}\right)\right] \\ =1-\cos \frac{A}{2}\left[\cos \left(\frac{\pi}{2}-\frac{B+C}{2}\right)-\sin \left(\frac{B-C}{2}\right)\right] \\ =1-\cos \frac{A}{2}\left[\sin \left(\frac{B+C}{2}\right)-\sin \left(\frac{B-C}{2}\right)\right] \\ =1-\cos \frac{A}{2} \cdot 2 \cos \left(\frac{\frac{B+C}{2}+\frac{B-C}{2}}{2}\right) \sin \left(\frac{\frac{B+C}{2}-\frac{B-C}{2}}{2}\right) \\ =1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}=R.H.S. 
Example:27. \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}-\cos ^2 \frac{C}{2} =2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}
Solution: \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}-\cos ^2 \frac{C}{2} =2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \\ \text { L.H.S. } \cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}-\cos ^2 \frac{C}{2} \\ =\cos ^2 \frac{A}{2}-\left[\cos ^2 \frac{C}{2}-\cos ^2 \frac{B}{2}\right] \\=\cos ^2 \frac{A}{2}-\sin \left(\frac{C}{2}+\frac{B}{2}\right) \sin \left(\frac{B}{2}-\frac{C}{2}\right) [ \sin (A+B) \sin (A-B)=\cos ^2 B-\cos ^2 A सूत्र से]
= \cos ^2 \frac{A}{2}-\sin \left(\frac{\pi}{2}-\frac{A}{2}\right) \sin \left(\frac{B-C}{2}\right) \\ =\cos ^2 \frac{A}{2}-\cos \frac{A}{2} \sin \left(\frac{B-C}{2}\right) \\ =\cos \frac{A}{2}\left[\cos \frac{A}{2}-\sin \left( \frac{B-C}{2}\right)\right] \\ =\cos \frac{A}{2}\left[\cos \left(\frac{\pi}{2}-\frac{B+C}{2}\right)-\sin \left( \frac{B-C}{2} \right)\right] \\ =\cos \left(\frac{A}{2}\right) \cdot\left[\sin \left(\frac{B+C}{2}\right)-\sin \left(\frac{B-C}{2}\right)\right] \\ =\cos \left(\frac{A}{2}\right) \cdot 2 \cos \left(\frac{\frac{B+C}{2}+\frac{B-C}{2}}{2}\right) \sin \left(\frac{\frac{B+C}{2}-\frac{B-C}{2}}{2}\right)
[ \sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right) सूत्र से]
=2 \cos \left(\frac{A}{2}\right) \cos \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)=R.H.S.
Example:28. \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2} =\cot \frac{A}{2} \cot \frac{B}{2} \cos \frac{C}{2}
Solution: A+B+C=180^{\circ} \\ \Rightarrow \quad \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^{\circ} \\ \tan \left(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\right)=\tan 90^{\circ} \\ \Rightarrow \frac{ \tan \frac{A}{2} +\tan \frac{B}{2}+\tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}-\tan \frac{B}{2} \tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{C}{2}}=\frac{1}{0} \left[\infty=\frac{1}{0}\right] \\ \Rightarrow 1-\tan \frac{A}{2} \tan \frac{B}{2}-\tan \frac{B}{2} \tan \frac{C}{2}-\tan \frac{A}{2} \tan \frac{C}{2}=0 \\ \Rightarrow \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{A}{2} \tan \frac{C}{2}=1
दोनों पक्षों में \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} का भाग देने परः
\Rightarrow \frac{\tan \frac{B}{2} \tan \frac{B}{2}}{\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}+ \frac{\tan \frac{B}{2} \tan \frac{C}{2}}{\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}} +\frac{\tan \frac{A}{2} \tan \frac{C}{2}}{\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}=\frac{1}{\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}} \\ \Rightarrow \frac{1}{\tan \frac{C}{2}}+\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{B}{2}}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} \\ \Rightarrow \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}
Example:29. \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right) +\tan \left(\frac{\pi+C}{4}\right)=\cot \left(\frac{B+C}{4}\right) \cot \left(\frac{C+A}{4}\right) \cot \left(\frac{A+B}{4}\right)
Solution: \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right) +\tan \left(\frac{\pi+C}{4}\right)=\cot \left(\frac{B+C}{4}\right) \cot \left(\frac{C+A}{4}\right) \cot \left(\frac{A+B}{4}\right) \\ A+B+C=\pi \\ \pi+A+\pi+B+\pi+C=3 \pi+\pi \\ \frac{\pi+A}{4}+\frac{\pi+B}{4}+\frac{\pi+C}{4}=\pi \\ \Rightarrow \tan \left[\left(\frac{\pi+A}{4}\right)+\left(\frac{\pi+B}{4}\right)+\left(\frac{\pi+C}{4}\right)\right]=\tan \pi \\ \Rightarrow \frac{\tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right)-\tan \left(\frac{\pi+A}{4}\right) \tan \left(\frac{\pi+B}{4}\right) \tan \left(\frac{\pi+C}{4}\right)}{1-\tan \left(\frac{\pi+A}{4}\right) \tan \left(\frac{\pi+B}{4}\right)-\tan \left(\frac{\pi+B}{4}\right) \tan \left(\frac{\pi+C}{4}\right)-\tan \left(\frac{\pi+A}{4}\right) \tan \left(\frac{\pi+C}{4}\right)}=0 \\ \Rightarrow \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right)-\tan \left(\frac{\pi+A}{4}\right) \tan \left(\frac{\pi+B}{4}\right) \tan \left(\frac{\pi+C}{4}\right)=0 \\ \Rightarrow \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right) =\tan \left(\frac{\pi+A}{4}\right) \tan \left(\frac{\pi+B}{4}\right) \tan \left(\frac{\pi+C}{4}\right) \\ \Rightarrow \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right) =\tan \left(\frac{2 \pi-\overline{B+C}}{4}\right) \tan \left(\frac{2 \pi-\overline{A+C}}{4}\right) \tan \left(\frac{2 \pi-\overline{A+B}}{4}\right) \\ \tan \left( \frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right) = \tan \left(\frac{\pi}{2}-\frac{\overline{B+C}}{4}\right) \tan \left(\frac{\pi}{2}-\frac{\overline{A+C}}{4}\right) \tan \left(\frac{\pi}{2}-\frac{\overline{A+B}}{4}\right) \\ \Rightarrow \tan \left(\frac{\pi+A}{4}\right)+\tan \left(\frac{\pi+B}{4}\right)+\tan \left(\frac{\pi+C}{4}\right) =\cot \left(\frac{B+C}{4}\right) \cot \left(\frac{A+C}{4}\right) \cot \left(\frac{A+B}{4}\right)
उपर्युक्त उदाहरणों के द्वारा त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Conditional Identities in Trigonometry),प्रतिबन्धित त्रिकोणमितीय सर्वसमिकाएँ (Conditional Trigonometrical Identities) को समझ सकते हैं।

3.त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ पर आधारित समस्याएँ (Problems Based on Conditional Identities in Trigonometry):

Conditional Identities in Trigonometry

(1.) A+B+C=\pi हो तो सिद्ध कीजिए:
\cos \frac{A}{2}+\cos \frac{B}{2}-\cos \frac{C}{2}=4 \cos \left(\frac{\pi+A}{4}\right) \cos \left(\frac{\pi+B}{4}\right) \cos \left(\frac{\pi-C}{4}\right)
(2.)यदि A+B+C=180° हो तो सिद्ध कीजिए
\sin ^2 A+\sin ^2 B+\sin ^2 C=2+2 \cos A \cos B \cos C
उपर्युक्त सवालों को हल करने पर त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Conditional Identities in Trigonometry),प्रतिबन्धित त्रिकोणमितीय सर्वसमिकाएँ (Conditional Trigonometrical Identities) को ठीक से समझ सकते हैं।

Also Read This Article:- Conditional Trigonometrical Identities

4.त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Frequently Asked Questions Related to Conditional Identities in Trigonometry),प्रतिबन्धित त्रिकोणमितीय सर्वसमिकाएँ (Conditional Trigonometrical Identities) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

त्रिकोणमिति में प्रतिबन्धित सर्वसमिकाएँ (Conditional Identities in Trigonometry) के इस
आर्टिकल में प्रतिबन्धित सर्वसमिका A+B+C=180° आदि के द्वारा त्रिकोणमितीय सवालों
को हल करके समझने का प्रयास करेंगे।

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