1.त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th),त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities):

Trigonometric Identities Class 10th

त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th) के इस आर्टिकल में कुछ विशिष्ट सवालों को हल करेंगे जो त्रिकोणमितीय सर्वसमिकाओं पर आधारित हैं।
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:- Application of Quadratic Equation 10th

2.त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं के उदाहरण (Trigonometric Identities Class 10th Examples):

निम्नलिखित सर्वसमिकाएँ सिद्ध कीजिए,जहाँ वे कोण,जिनके लिए व्यंजक परिभाषित है,न्यून कोण है:
Example:1. \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta
Solution: \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta \\ \text{L.H.S. } \frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta} \\ =\frac{(1+\sin \theta)^2+\cos ^2 \theta}{\cos \theta(1+\sin \theta)} \\ =\frac{1+2 \sin \theta+\sin ^2 \theta+\cos ^2 \theta}{\cos \theta(1+ \sin \theta)} \\ =\frac{1+2 \sin \theta+1}{\cos \theta(1+\sin \theta)}\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\ =\frac{2(1+\sin \theta)}{\cos \theta(1+\sin \theta)} \\ =\frac{2}{\cos \theta}=2 \sec \theta=\text { R.H.S. }
Example:2. \frac{\sin ^4 \theta-\cos ^4 \theta}{\sin ^2 \theta-\cos ^2 \theta}=1
Solution: \frac{\sin ^4 \theta-\cos ^4 \theta}{\sin ^2 \theta-\cos ^2 \theta}=1 \\ \text { L.H.S. } \frac{\sin ^4 \theta-\cos ^4 \theta}{\sin ^2 \theta-\cos ^2 \theta} \\ =\frac{\left(\sin ^2 \theta-\cos ^2 \theta\right)\left(\sin ^2 \theta+\cos ^2 \theta\right)}{\sin ^2 \theta-\cos ^2 \theta} \\ =\sin ^2 \theta+\cos ^2 \theta=1=\text { R.H.S. } 
Example:3. \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec^2 \theta
Solution: \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta \\ \text { L:H.S } \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta} \\ \theta =\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)} \\ =\frac{2}{1-\sin ^2 \theta} \\ =\frac{2}{\cos ^2 \theta} \quad\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\ =2 \sec ^2 \theta=\text { R.H.S }
Example:4. (\sec \theta-\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta
Solution: (\sec \theta-\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta \\ \text { L.H.S }(\sec \theta-\cos \theta)(\cot \theta+\tan \theta) \\ =\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right)\left[ \sec \theta=\frac{1}{\cos \theta} \text{ तथा  } \tan \theta =\frac{\sin \theta}{\cos \theta} \right]\\ =\left(\frac{1-\cos ^2 \theta}{\cos \theta}\right) \left(\frac{\cos ^2 \theta+\sin \theta}{\cos \theta}\right) \\ =\frac{\sin ^2 \theta}{\cos ^2 \theta} \times \frac{1}{\sin \theta \cos \theta} \\ =\frac{\sin \theta}{\cos ^2 \theta}=\tan \theta \sec \theta =\text { R.H.S }
Example:5. \cot \theta-\tan \theta=\frac{1-2 \sin ^2 \theta}{\cos \theta \sin \theta}
Solution: \cot \theta-\tan \theta=\frac{1-2 \sin ^2 \theta}{\cos \theta \sin \theta} \\ \text{R.H.S. } \frac{1-2 \sin ^2 \theta}{\cos \theta \sin \theta} \\ =\frac{1-\sin ^2 \theta-\sin ^2 \theta}{\cos \theta \sin \theta} \\ =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos ^2 \theta \cdot \sin \theta} \\ =\frac{\cos ^2 \theta}{\cos \theta \sin \theta}-\frac{\sin ^2 \theta}{\cos \theta \sin \theta} \\ =\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \\=\cot \theta-\tan \theta=\text{L.H.S.}
Example:6. \operatorname{cosec}^6 \theta-\cot ^6 \theta=1+3 \operatorname{cosec}^2 \theta \cot ^2 \theta
Solution: \operatorname{cosec}^6 \theta-\cos ^6 \theta=1+3 \operatorname{cosec}^2 \theta \cot \theta \\ \text { L.H.S. } \operatorname{cosec}^6 \theta-\cot ^6 \theta \\ =\left(\operatorname{cosec}^2 \theta\right)^3 -\left(\cot ^2 \theta\right)^3 \\ =\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right) \left[\operatorname{cosec}^4 \theta+\cot ^4 \theta+ \operatorname{cosec}^2 \theta \cot ^2 \theta\right] \\ =(1)\left[\operatorname{cosec}^4 \theta+\cot ^4 \theta-2 \operatorname{cosec}^2 \theta \cot ^2 \theta+3 \operatorname{cosec}^2 \theta \cot ^2 \theta\right] \\ \left[\because cosec^2 \theta-\cot^2 \theta=1\right] \\ =\left[\left(\operatorname{cosec}^2 \theta-\cot^2 \theta\right)^2+3 \operatorname{cosec}^2 \theta \cot^2 \theta\right] \\=\left[1^2+3 \operatorname{cosec}^2 \theta \cot ^2 \theta\right] \\ =1+3 \operatorname{cosec}^2 \theta \cot ^2 \theta=\text { R.H.S. }
विकल्पतः \text { R.H.S. } 1+3\operatorname{cosec}^2 \theta \cot ^2 \theta \\ =\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)^2+3 \operatorname{cosec}^2 \theta \cot ^2 \theta \\ =\operatorname{cosec}^4 \theta+\cot ^4 \theta-2 \operatorname{cosec}^2 \theta \cot ^2 \theta +3 \operatorname{cosec}^2 \theta \cot ^2 \theta \\ =(1)\left(\operatorname{cosec}^4 \theta+\cot ^4 \theta+\operatorname{cosec}^2 \theta \cot ^2 \theta\right) \\ =\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)\left[\left(\operatorname{cosec}^2 \theta\right)^2+\left(\cot ^2 \theta\right)^2+\operatorname{cosec}^2 \theta \cot ^2 \theta\right] \\ =\left(\operatorname{cosec}^2 \theta\right)^3-\left(\cot ^2 \theta\right)^3 \\ =\operatorname{cosec}^6 \theta-\cot ^6 \theta=\text { L.H.S. }
Example:7. \sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta) =\operatorname{cosec} \theta+\sec \theta
Solution: \sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)=\frac{\operatorname{cosec} \theta+1}{\sec \theta} \\ \text { L.H.S. } \sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta) \\ =\sin \theta \left(1+\frac{\sin \theta}{\cos \theta}\right)+\cos \theta\left(1+\frac{\cos \theta}{\sin \theta}\right)\left[ \cot \theta=\frac{\cos \theta}{\sin \theta} \right] \\ =\sin \theta\left(\frac{\cos \theta+\sin \theta}{\cos \theta}\right)+\cos \theta\left(\frac{\sin \theta+\cos \theta}{\sin \theta}\right) \\ =(\cos \theta+\sin \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right) \\ =(\cos \theta+\sin \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos ^2 \theta \sin \theta}\right) \\ =(\cos \theta+\sin \theta) \times \frac{1}{\cos \theta \sin \theta} \\ =\frac{\cos \theta}{\cos \theta \sin \theta}+ \frac{\sin \theta}{\cos \theta \sin \theta} \\ =\frac{1}{\sin \theta} +\frac{1}{\cos \theta}=\operatorname{cosec} \theta+\sec \theta=\text { R.H.S. }
विकल्पतः \text { L.H.S. }=\frac{\cos \theta+\sin \theta}{\cos \theta \sin \theta} \\ =\operatorname{cosec} \theta+\sec \theta \\ =\frac{1}{\sin \theta}+\frac{1}{\cos \theta} \\ =\frac{\cos \theta+\sin \theta}{\sin \theta \cos \theta} \\ \text { L.H.S}= \text{R.H.S } 
Example:8. \frac{\sin ^3 \theta+\cos ^3 \theta}{(\sin \theta+\cos \theta)}=1-\cos \theta \sin \theta
Solution: \frac{\sin ^3 \theta+\cos ^3 \theta}{(\sin \theta+\cos \theta)}=1-\cos \theta \sin \theta \\ \text{L.H.S. }=\frac{\sin ^3 \theta+\cos ^3 \theta}{(\sin \theta+\cos \theta)} \\ =\frac{(\sin \theta+\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta-\cos \theta \sin \theta\right)}{(\sin \theta+\cos \theta)} \left[ \because a^3+b^3=(a+b)(a^2+b^2-ab) \right]\\ =\sin ^2 \theta+\cos ^2 \theta-\sin \theta \cos \theta\\ =1-\sin \theta \cos \theta=\text { R.H.S. }
विकल्पतः 1-\cos \theta \sin \theta\\ \sin ^2 \theta+\cos ^2 \theta-\cos \theta \sin \theta \\ =\frac{(\sin \theta+\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta-\cos \theta \sin \theta\right)}{(\sin \theta+\cos \theta)} \\ =\frac{\sin ^3+\cos ^3 \theta}{(\sin \theta+\cos \theta)}=\text { L.H.S. }
Example:9. \frac{\cot \theta}{1+\tan \theta}=\frac{\cot \theta-1}{2-\sec ^2 \theta}
Solution: \frac{\cot \theta}{1+\tan \theta}=\frac{\cot \theta-1}{2-\sec ^2 \theta} \\ \text{Method 1} \\ \text { L.H.S. } \frac{\cot \theta}{1+\tan \theta}
अंश व हर को  1-\tan \theta से गुणा करने पर:
\frac{\cot \theta (1-\tan \theta)}{(1+\tan \theta)(1-\tan \theta)} \\ =\frac{\cot \theta-\cot \theta \tan \theta}{1-\tan ^2 \theta} \\ =\frac{\cot \theta-1}{1-(\sec ^2 \theta-1)}
[ \because \cot \theta \tan \theta=1 तथा 1+ \tan^2 \theta= \sec^2 \theta ]
=\frac{\cot \theta-1}{1-\sec ^2 \theta+1}\\ =\frac{\cot \theta-1}{2-\sec ^2 \theta}=\text { R.H.S. } \\ \text { Method 2} \\ \text { R.H.S. } =\frac{\cot \theta-1}{2-\sec ^2 \theta} \\ =\frac{\cot \theta-\cot \theta \tan \theta }{1+1-\sec ^2 \theta} \\ =\frac{\cot \theta\left(1-\tan \theta\right)}{1-\left(\sec ^2 \theta-1\right)} \left[ \because 1=\cot \theta \tan \theta \right] \\ =\frac{\cot \theta(1-\tan \theta)}{1-\tan^2 \theta} \left[\because \sec ^2 \theta-1=\tan ^2 \theta \right] \\ =\frac{\cot \theta(1-\tan \theta)}{(1+\tan \theta)(1-\tan \theta)} \\ =\frac{\cot \theta}{1+\tan \theta}=\text { L.H.S. } \\ \text { Method : 3 } \frac{ \cot \theta}{1+\tan \theta} \\ =\frac{\frac{\cos \theta}{\sin \theta}}{1+\frac{\sin \theta}{\cos \theta}} \\ =\frac{\cos ^2 \theta}{\sin \theta( \sin \theta+\cos \theta)} \cdots(1) \\ \text{R.H.S. } \frac{\cot \theta-1}{2-\cos^2 \theta} \\ =\frac{\frac{\cos \theta}{\sin \theta}-1}{2-\frac{1}{\cos ^2 \theta}} \\ =\frac{(\cos \theta-\sin \theta) \cos ^2 \theta}{\sin \theta\left(2 \cos ^2 \theta-1\right)} \\ =\frac{(\cos \theta-\sin \theta) \cos ^2 \theta}{\sin \theta\left(\cos ^2 \theta+\cos ^2 \theta-1\right)} \\ =\frac{(\cos \theta-\sin \theta) \cos ^2 \theta}{\sin \theta\left[\cos ^2 \theta-\left(1-\cos ^2 \theta\right)\right]} \\ =\frac{(\cos \theta-\sin \theta) \cos ^2 \theta}{\sin \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)} \\ =\frac{(\cos \theta-\sin \theta) \cos ^2 \theta}{\sin \theta(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)} \\ =\frac{\cos ^2 \theta}{\sin \theta(\cos \theta+\sin \theta)} \cdots(2)
(1) व (2) सेः
L.H.S.=R.H.S.
Example:10.  \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin \theta}=\operatorname{cosec} \theta+\cot \theta
Solution:  \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin \theta}= \operatorname{cosec} \theta+\cot \theta \\ \text { L.H.S. } \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} =\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}} \\ =\sqrt{\frac{(1 +\cos \theta)^2}{1-\cos ^2 \theta}} \\ =\frac{1+\cos \theta}{\sqrt{\sin ^2 \theta}} \\ =\frac{1+\cos \theta}{\sin \theta}=\text{M.H.S.} \\ =\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} \\ =\operatorname{cosec} \theta+\cot \theta=\text{R.H.S}

Trigonometric Identities Class 10th

Example:11. \left(1+\cot ^2 \theta\right)(1-\cos \theta)(1+\cos \theta)=1
Solution: \left(1+\cot ^2 \theta\right)(1-\cos \theta)(1+\cos \theta)=1 \\ \text { L.H.S. }\left(1+\cot ^2 \theta\right)(1-\cos \theta)(1+\cos \theta) \\ =\left(\operatorname{cosec}^2 \theta\right)\left(1-\cos ^2 \theta\right)\left[\because 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta\right] \\=\operatorname{cosec}^2 \theta \sin ^2 \theta \\ =1=\text { R.H.S. }[ \because \operatorname{cosec} \theta \sin \theta=1]
विकल्पतः
\text { R.H.S. } =1 \\ =1^2 \\ \\=(\operatorname{cosec} \theta \sin \theta)^2[\because \operatorname{cosec} \theta \sin \theta=1] \\ =\operatorname{cosec}^2 \theta \sin ^2 \theta \\ =\left(1+\cot ^2 \theta\right)\left(1-\cos ^2 \theta\right) \\ =\left(1+\cot ^2 \theta\right)(1-\cos \theta)(1+\cos \theta)=\text{L.H.S.}
Example:12. \frac{1}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta}=\frac{1}{\cos \theta}-\frac{1}{\sec \theta+\tan \theta}
Solution: \frac{1}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta}=\frac{1}{\cos \theta}-\frac{1}{\sec \theta+\tan \theta} \\ \text { L.H.S.} \frac{1}{\sec \theta-\tan \theta}-\frac{1}{\cos \theta} \\=\frac{(\sec \theta+\tan \theta)}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}-\frac{1}{\cos \theta} \\ =\frac{\sec \theta+\tan \theta}{\sec ^2 \theta-\tan ^2 \theta}-\frac{1}{\cos \theta} \\ =\sec \theta+\tan \theta-\sec \theta\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right] \\ =\frac{1}{\cos \theta}-(\sec \theta-\tan \theta) \\ =\frac{1}{\cos \theta}-\frac{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}{(\sec \theta+\tan \theta)} \\ =\frac{1}{\cos \theta}-\frac{\sec ^2 \theta-\tan ^2 \theta}{\sec \theta+\tan \theta} \\ =\frac{1}{\cos \theta}=\frac{1}{\sec \theta+\tan \theta}=\text { R.H.S. }
Example:13. \frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=1-2 \sec \theta \tan \theta
Solution: \frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=1-2 \sec \theta \tan \theta+2 \tan^2 \theta \\ \text { L.H.S. } \frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}
अंश व हर को \sec \theta \tan \theta से गुणा करने परः
= \frac{(\sec \theta-\tan \theta)(\sec \theta-\operatorname{sen} \theta)}{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)} \\ =\frac{\sec ^2 \theta+\tan ^2 \theta-2 \sec \theta \tan \theta}{\sec ^2 \theta-\tan ^2 \theta} \\=1+\tan ^2 \theta+\tan ^2 \theta-2 \sec \theta \tan \theta  \\ \left[ \because \sec ^2 \theta =1+\tan ^2 \theta \text{ तथा } \sec ^2 \theta-\tan ^2 \theta=1 \right] \\ =1- 2\sec \theta \cdot \tan \theta+2 \tan ^2 \theta=\text { R.H.S. }
Example:14. \sin ^2 \theta \cos \theta+\tan \theta \sin \theta+\cos ^3 \theta=\sec \theta
Solution: \sin ^2 \theta \cos \theta+\tan \theta \sin \theta+\cos ^3 \theta=\sec \theta \\ \text { L.H.S. } \sin ^2 \theta \cos \theta+\tan \theta \sin \theta+\cos ^3 \theta \\ = \sin ^2 \theta \cos \theta+\cos ^3 \theta+\tan \theta \sin \theta \\ =\cos \theta\left(\sin ^2 \theta+\cos ^2 \theta\right)+\tan \theta \sin \theta \\ =\cos \theta +\frac{\sin \theta}{\cos \theta} \cdot \sin \theta \left[ \because \tan \theta=\frac{\sin \theta}{\cos \theta}\right] \\ =\cos \theta+\frac{\sin ^2 \theta}{\cos \theta} \\ =\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos \theta}=\frac{1}{\cos \theta}=\sec \theta=\text{ R.H.S.}
Example:15. (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)=2
Solution: (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)=2 \\ \text{L.H.S. } (1+\cot \theta -\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta) \\ =\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left[ \cot \theta=\frac{\cos \theta}{\sin \theta} ,\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right] \\ =\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right)\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right) \\ =\frac{(\sin \theta+\cos \theta)^2-1}{\sin \theta \cos \theta} \quad\left[ \because (a+b)(a-b)=a^2-b^2\right] \\ =\frac{\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\ =\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\ =\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2=\text { R.H.S. }
Example:16. \frac{\cot \theta \cos \theta}{\cot \theta+\cos \theta}=\frac{\cot \theta-\cos \theta}{\cot \theta \cos \theta}
Solution:\frac{\cot \theta \cos \theta}{\cot \theta+\cos \theta}=\frac{\cot \theta-\cos \theta}{\cot \theta \cos \theta} \\ \text{L.H.S.} \frac{\cot \theta \cos \theta}{\cos t \theta+\cos \theta}
अंश व हर को \cot \theta-\cos \theta से गुणा करने परः
\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{(\cot \theta+\cos \theta)(\cot \theta-\cos \theta)} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\cot ^2 \theta-\cos ^2 \theta} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\operatorname{cosec}^2 \theta-1-\cos ^2 \theta} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\operatorname{cosec}^2 \theta-1-\cos ^2 \theta} \left[ \because \cot^2 \theta=\operatorname{cosec}^2 \theta-1 \right]\\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\frac{1}{\sin ^2 \theta}-1-\cos ^2 \theta } \left[ \because \operatorname{cosec}^2 \theta=\frac{1}{\sin ^2 \theta} \right] \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\frac{1-\sin ^2 \theta-\cos ^2 \theta \sin ^2 \theta}{\sin^2 \theta}}\\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\operatorname{cosec}^2 \theta\left(\cos ^2 \theta-\cos ^2 \theta \sin ^2 \theta\right)} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{ \operatorname{cosec}^2 \theta \cdot \cos ^2 \theta\left(1-\sin ^2 \theta\right)} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\operatorname{cosec}^2 \theta \cos ^2 \theta \cos ^2 \theta} \\ =\frac{(\cot \theta \cos \theta)(\cot \theta-\cos \theta)}{\cot ^2 \theta \cos ^2 \theta}\left[ \because \operatorname{cosec}^2 \theta \cos ^2 \theta =\cot ^2 \theta \right] \\=\frac{\cot \theta-\cos \theta}{\cot \theta \cos \theta}=\text { R.H.S }
Example:17. \frac{\tan ^3 \theta-1}{\tan ^2 \theta-1}=\sec ^2 \theta+\tan \theta
Solution: \frac{\tan ^3 \theta-1}{\tan \theta-1}=\sec ^2 \theta+\tan \theta \\ \text { L.H.S. } \frac{\tan ^3 \theta-1}{\tan \theta-1} \\ =\frac{(\tan \theta-1)\left(\tan ^2 \theta+1+\tan \theta\right)}{(\tan \theta-1)} \\ =1+\tan ^2 \theta+\tan \theta \\ =\sec ^2 \theta+\tan \theta=\text { R.H.S. }
Example:18. \sec ^6 \theta-\tan ^6 \theta=1+3 \tan ^2 \theta+3 \tan 4 \theta 
Solution: \sec ^6 \theta-\tan ^6 \theta=1+3 \tan ^2 \theta+3 \tan ^2 \theta \\ \text { L.H.S. } \sec ^6 \theta-\tan ^6 \theta \\ =\left(\sec ^2 \theta\right)^3-\left(\tan ^2 \theta\right)^3 \\ =\left(\sec ^2 \theta-\tan ^2 \theta\right)\left[\sec ^4 \theta+\tan ^4 \theta+\sec ^2 \theta \tan ^2 \theta\right] \\ =(1)\left[\sec ^4 \theta+\tan ^4 \theta-2 \sec ^2 \theta \tan ^2 \theta+3 \sec ^2 \theta \tan ^2 \theta\right] \\ =\left[\left(\sec ^2 \theta-\tan ^2 \theta\right)^2+3 \sec ^2 \theta \tan ^2 \theta\right] \\ =\left[1^2+3 \sec ^2 \theta \tan ^2 \theta\right]\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right] \\ =1+3 \sec ^2 \theta \tan ^2 \theta=\text{R.H.S.}
Example:19. \tan ^2 A-\tan ^2 B=\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 A \cos ^2 B}=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cos ^2 B}
Solution: \tan ^2 A-\tan ^2 B=\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 A \cos ^2 B}=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cos ^2 B}\\ \text { L.H.S.} \tan ^2 A -\tan ^2 B \\ =\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B} \\ =\frac{\sin ^2 A\left(\cos ^2 B\right)-\cos ^2 A \sin ^2 B}{\cos ^2 A \cos ^2 B} \\ =\frac{\left(1-\cos ^2 A\right) \cos ^2 B-\cos ^2 A\left(1-\cos ^2 B\right)}{\cos ^2 A \cos ^2 B} \\ =\frac{\cos ^2 B-\cos ^2 A \cos ^2 B-\cos ^2 A+\cos ^2 A \cos ^2 B}{\cos ^2 A \cos ^2 B} \\ =\frac{\cos ^2 B-\cos ^2 A}{\cos ^2 A \cos ^2 B}=\text{M.H.S.} \\=\frac{1-\sin ^2 B-\left(1-\sin ^2 A\right)}{\cos ^2 A \cos ^2 B} \\ =\frac{1-\sin ^2 B-1+\sin ^2 A}{\cos ^2 A \cos ^2 B} \\ =\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}=\text { R.H.S. }
Example:20. \frac{\tan ^2 \theta}{\tan ^2 \theta-1}+\frac{\operatorname{cosec}^2 \theta}{\sec ^2 \theta-\operatorname{cosec}^2 \theta}=\frac{1}{\sin ^2 \theta-\cos ^2 \theta}
Solution: \frac{\tan ^2 \theta}{\tan ^2 \theta-1}+\frac{\operatorname{cosec}^2 \theta}{\sec ^2 \theta-\operatorname{cosec}^2 \theta}=\frac{1}{\sin ^2 \theta-\cos ^2 \theta} \\ \text { L.H.S } \frac{\tan ^2 \theta}{\tan ^2 \theta-1}+\frac{\operatorname{cosec}^2 \theta}{\sec ^2 \theta-cosec^2 \theta} \\ =\frac{\frac{\sin ^2 \theta}{\cos ^2 \theta}}{\frac{\sin ^2 \theta}{\cos ^2 \theta}-1}+\frac{\frac{1}{\sin ^2 \theta}}{\frac{1}{\cos ^2 \theta}-\frac{1}{\sin ^2 \theta}} \\ =\frac{\sin ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta}+\frac{\cos ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta} \\ =\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta-\cos ^2 \theta} \\ =\frac{1}{\sin ^2 \theta-\cos ^2 \theta}=\text { R.H.S. }
Example:21. \frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta-1}+ \frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta+1}=2 \sec ^2 \theta
Solution: \frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta-1}+\frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta+1}=2 \sec ^2 \theta \\ \text{L.H.S. } \frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta-1}+\frac{\operatorname{cosec} \theta}{\operatorname{cosec} \theta+1} \\ =\frac{\operatorname{cosec} \theta(\operatorname{cosec} \theta+1)+\operatorname{cosec} \theta(\operatorname{cosec} \theta-1)}{(\operatorname{cosec} \theta-1)(\operatorname{cosec} \theta+1)} \\ =\frac{\operatorname{cosec}^2 \theta+\operatorname{cosec} \theta+\operatorname{cosec}^2 \theta-\operatorname{cosec} \theta}{\operatorname{cosec}^2 \theta-1} \\ =\frac{2 \operatorname{cosec}^2 \theta}{\cot ^2 \theta} \\ =\frac{\frac{2}{\sin ^2 \theta}}{\frac{\cos ^2 \theta}{\sin ^2 \theta}} \\ =\frac{2}{\sin ^2 \theta} \times \frac{\sin ^2 \theta}{\cos ^2 \theta} \\ =\frac{2}{\cos ^2 \theta}=2 \sec ^2 \theta=\text { R.H.S. }
Example:22. (\tan A+\operatorname{cosec} B)^2-(\cot B-\sec A)^2 =2 \tan A \cot B(\operatorname{cosec} A+\sec B)
Solution: (\tan A+\operatorname{cosec} B)^2-(\cot B-\sec A)^2 \\ =2 \tan A \cot B(\operatorname{cosec} A+\sec B) \\ \text { L.H.S }(\tan A+ \operatorname{cosec} B)^2-(\cot B-\sec A)^2 \\ =\tan ^2 A+ \operatorname{cosec}^2 B-\left(\cot ^2 B+\sec ^2 A +2 \tan A \operatorname{cosec} B -2 \cot B \cdot \sec A\right) \\ =\tan ^2 A+\operatorname{cosec}^2 B+2 \tan A \operatorname{cosec} B -\cot ^2 B-\sec ^2 A+2 \cot B \sec A \\ =-\left(\sec ^2 A-\tan ^2 A\right)+\left(\operatorname{cosec}^2 B-\cot ^2 B\right) +2 \tan A \cot B\left( \frac{\operatorname{cosec} B}{\cot B}+\frac{\sec A}{\tan A}\right) \\ =-1+1+2 \tan A \cot B(\sec B+\operatorname{cosec} A) \\ =2 \tan A \cot B(\sec B+\operatorname{cosec} A)=\text{R.H.S.}
Example:23. 2 \sec ^2 \theta-\sec ^4 \theta-2 \operatorname{cosec}^2 \theta+ \operatorname{cosec}^4 \theta=\cot ^4 \theta-\tan ^4 \theta
Solution: 2 \sec ^2 \theta-\sec ^4 \theta-2 \operatorname{cosec}^2 \theta+ \operatorname{cosec}^4 \theta=\cot ^4 \theta-\tan ^4 \theta \\ \text { L.H.S. } 2 \sec ^2 \theta-\sec ^4 \theta-2 \operatorname{cosec}^2 \theta+\operatorname{cosec}^4 \theta \\ =2\left(\sec ^2 \theta-\operatorname{cosec}^2 \theta\right)-\left(\sec ^4 \theta-\operatorname{cosec}^4 \theta\right) \\ =2\left(\sec ^2 \theta-\operatorname{cosec}^2 \theta\right)-\left(\sec ^2 \theta-\operatorname{cosec}^2 \theta\right) \left(\sec ^2 \theta+\operatorname{cosec}^2 \theta\right) \\ =\left(\sec ^2 \theta-\operatorname{cosec}^2 \theta\right)\left(2-\sec ^2 \theta -\operatorname{cosec}^2 \theta\right) \\ =\left(\tan ^2 \theta-\cot ^2 \theta\right)\left(2-1-\tan ^2 \theta-1-\cot ^2 \theta\right) \\ =\left(\cot ^2 \theta-\tan ^2 \theta\right)\left(\cot ^2 \theta+\tan ^2 \theta\right) \\ =\cot ^4 \theta-\tan ^4 \theta =\text { R.H.S. }
Example:24.यदि \sin \theta+\cos \theta=p और \sec \theta+\operatorname{cosec} \theta=q ,तो सिद्ध कीजिए q\left(p^2-1\right)=2 p 
Solution: q\left(p^2-1\right)=2 p \\ \text{L.H.S. } q\left(p^2-1\right) \\=(\sec \theta+\operatorname{cosec} \theta)\left[(\sin \theta+\cos \theta)^2-1\right] \\ =\frac{1}{\cos \theta+\sin \theta)} \left(\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1\right) \\ =\frac{(\sin \theta+\cos \theta)}{\sin \theta \cos \theta}(1+2 \sin \theta \cos \theta-1) \\ =\frac{(\sin \theta+\cos \theta)}{\sin \theta \cos \theta} \times 2 \sin \theta \cos \theta \\ =2(\sin \theta+\cos \theta) \\ =2 p= \text{R.H.S.}
उपर्युक्त उदाहरणों के द्वारा त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th),त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities) को समझ सकते हैं।

3.त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं के सवाल (Trigonometric Identities Class 10th Questions):

Trigonometric Identities Class 10th

निम्नलिखित सर्वसमिकाओं को सिद्ध कीजिए:

(1.) \tan \theta+\cot \theta=\operatorname{cosec} \theta \sec \theta
(2.) \operatorname{cosec}^2 \theta+\sec ^2 \theta=\operatorname{cosec}^2 \theta \sec ^2 \theta
(3.) \left(1+\tan ^2 \theta\right)(1+\sin \theta)(1-\sin \theta) =1
उपर्युक्त सवालों को हल करने पर त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th),त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities) को ठीक से समझ सकते हैं।

Also Read This Article:- Imp Illustration of Quadratic Equation

4.त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Frequently Asked Questions Related to Trigonometric Identities Class 10th),त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities) से सम्बन्धित अक्सर पूछे जाने वाले प्रश्न:

प्रश्न:1.त्रिकोणमितीय सर्वसमिकाएँ लिखिए। (Write Trigonometric Identities):

उत्तर: (1.)\sin ^2 \theta+\cos ^2 \theta=1 ,\sin^2 \theta=1- \cos ^2 \theta , \cos ^2 \theta=1-\sin ^2 \theta
(2.) 1+\tan^2 \theta=\sec^2 \theta, \sec ^2 \theta-\tan ^2 \theta=1, \sec ^2 \theta-1=\tan^2 \theta
(3.) 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta, \operatorname{cosec}^2 \theta-\cot ^2 \theta=1 ,\cot ^2 \theta=\operatorname{cosec}^2 \theta-1

प्रश्न:2.त्रिकोणमितीय अनुपात लिखिए। (Write Trigonometric Ratios):

उत्तर: \sin \theta=\frac{\text{लम्ब}}{\text{कर्ण}} , \cos \theta= \frac{\text{आधार}}{\text{कर्ण}}, \tan \theta=\frac{\text{लम्ब}}{\text{आधार}},\cot \theta= \frac{\text{आधार}}{\text{लम्ब}} , \sec \theta=\frac{\text{कर्ण}}{\text{आधार}} ,\operatorname{cosec} \theta= \frac{\text{कर्ण}}{\text{लम्ब}}

प्रश्न:3.त्रिकोणमितीय अनुपातों के मध्य क्या मूलभूत सम्बन्ध है? (What is the Relations Between Trigonometric Ratios?):

उत्तर: (1.) \sin \theta=\frac{1}{\operatorname{cosec} \theta} , \operatorname{cosec} \theta=\frac{1}{\sin \theta} , \sin \theta \operatorname{cosec} \theta=1
(2.) \cos \theta=\frac{1}{\sec \theta}, \sec \theta=\frac{1}{\cos \theta} , \cos \theta \sec \theta=1
(3.) \tan \theta=\frac{1}{\cot \theta}, \cot \theta=\frac{1}{\tan \theta}, \tan \theta \cot \theta=1
(4.) \tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}
उपर्युक्त प्रश्नों के उत्तर द्वारा त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th),त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities) के बारे में और अधिक जानकारी प्राप्त कर सकते हैं।

त्रिकोणमितीय सर्वसमिकाएँ कक्षा 10वीं (Trigonometric Identities Class 10th) के इस
आर्टिकल में कुछ विशिष्ट सवालों को हल करेंगे जो त्रिकोणमितीय सर्वसमिकाओं पर आधारित हैं।