1. Prove that $ \displaystyle \frac{{\tan (A+B)}}{{\cot (A-B)}}=\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$.
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| Solution $ \displaystyle \frac{{\tan (A+B)}}{{\cot (A-B)}}=\tan (A+B)\tan (A-B)$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin (A+B)}}{{\cos (A+B)}}\frac{{\sin (A-B)}}{{\cos (A-B)}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sin A\cos B+\cos A\sin B}}{{\cos A\cos B-\sin A\sin B}}\times \frac{{\sin A\cos B-\cos A\sin B}}{{\cos A\cos B+\sin A\sin B}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\sin }}^{2}}A{{{\cos }}^{2}}B-{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A{{{\cos }}^{2}}B-{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\sin }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\sin }}^{2}}A){{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A(1-{{{\sin }}^{2}}B)-(1-{{{\cos }}^{2}}A){{{\sin }}^{2}}B}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\sin }}^{2}}A{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\cos }}^{2}}A{{{\sin }}^{2}}B-{{{\sin }}^{2}}B+{{{\cos }}^{2}}A{{{\sin }}^{2}}B}}$ $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\sin }}^{2}}A-{{{\sin }}^{2}}B}}{{{{{\cos }}^{2}}A-{{{\sin }}^{2}}B}}$ |
2. Prove that $ \displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\frac{{\sin 2\theta }}{{\cos \theta }}=\operatorname{cosec}\theta $
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| Solution $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \displaystyle \frac{{\cos 2\theta }}{{\sin \theta }}+\displaystyle \frac{{\sin 2\theta }}{{\cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta }}{{\sin \theta \cos \theta }}+\displaystyle \frac{{\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos 2\theta \cos \theta +\sin 2\theta \sin \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \left( {2\theta -\theta } \right)}}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{{\cos \theta }}{{\sin \theta \cos \theta }}\\\\=\ \ \ \displaystyle \frac{1}{{\sin \theta }}\\\\=\ \ \ \operatorname{cosec}\theta \end{array}$ |
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