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।