| 1. \[{{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta \,{{\sin }^{2}}\theta .\] |
| 2. \[\left( \text{cosec }\theta -\sin \theta \right)\,\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1\] |
| Which of the above is / are correct? |
A) only is the identity
B) 2 only is the identity
C) Both 1 and 2 are the identities
D) Neither 1 nor 2 is the identity
Correct Answer: C
Solution :
\[{{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta .\,{{\sin }^{2}}\theta \] L.H.S \[={{\tan }^{2}}\theta -{{\sin }^{2}}\theta \] \[=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin }^{3}}\theta ={{\sin }^{2}}\theta \left( \frac{1}{{{\cos }^{2}}\theta }-1 \right)\] \[={{\sin }^{2}}\theta \left( \frac{1-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta } \right)=\frac{{{\sin }^{2}}\theta .{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\] \[=\left( \frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)\,{{\sin }^{2}}\theta ={{\tan }^{2}}\theta .{{\sin }^{2}}\theta \] = R.H.S So, it is an identity (ii) \[\left( \cos ec\,\theta -\sin \theta \right)\,\left( \sec \theta -\cos \theta \right)\] \[\left( \tan \theta +\cot \theta \right)=1\] L.H.S \[=\left( \frac{1}{\sin \theta }-\sin \theta \right)\left( \frac{1}{\cos \theta }-\cos \theta \right)\left( \frac{\sin \theta }{\cos \theta }+\frac{cos\theta }{\sin \theta } \right)\] \[=\left( \frac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \frac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\cos \theta .\sin \theta } \right)\] \[=\frac{{{\cos }^{2}}\theta }{\sin \theta }\times \frac{{{\sin }^{2}}\theta }{\cos \theta }\times \frac{1}{\cos \theta .\sin \theta }\] = 1 =R.H.S So, it is also as identity. \[\therefore \] Both 1 and 2 are the identities
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