A) Only I is the identity
B) Only II is the identity
C) Both I and II are the identities
D) Neither I nor II is the identity
Correct Answer: D
Solution :
| I. \[{{\tan }^{2}}\theta -{{\sin }^{2}}\theta =\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{2}}\theta ,\ne (2n+1)\frac{\pi }{2}\] |
| \[=\frac{{{\sin }^{2}}\theta \,\,(1-{{\cos }^{2}}\theta )}{{{\cos }^{2}}\theta }\theta \ne (2n+1)\frac{\pi }{2}\] |
| \[=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }{{\sin }^{2}}\theta ,\]\[\theta \ne \,\,(2n+1)\frac{\pi }{2}\] |
| \[={{\tan }^{2}}\theta {{\sin }^{2}}\theta ,\]\[\theta \ne (2n+1)\frac{\pi }{2}\] |
| II. \[(\text{cosec}\theta -\sin \theta )(\sec \theta -\cos \theta )(tan\theta +cot\theta )\] |
| \[=\left( \frac{1}{\sin \theta }-\sin \theta \right)\left( \frac{1}{\cos \theta }-\cos \theta \right)\left( \tan \theta +\frac{1}{\tan \theta } \right)\]\[[\because \theta \ne n\pi \,\,(2n+1)\frac{\pi }{2}]\] |
| \[=\frac{{{\cos }^{2}}\theta \cdot {{\sin }^{2}}\theta \cdot {{\sec }^{2}}\theta }{\sin \theta \cos \theta \tan \theta },\theta \ne n\pi \,\,(2n+1)\frac{\pi }{2}\] |
| \[=\sin \theta \cos \theta \frac{1}{{{\cos }^{2}}\theta }\cdot \frac{\cos \theta }{\sin \theta },\]\[\theta \ne n\pi ,\]\[(2n+1)\frac{\pi }{2}=1\] |
| Since, to become an identity, both statements Must be satisfied for every value of\[\theta \]. |
| Therefore, neither I nor II are the identity. |
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