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If \(\tan \theta = \frac{p}{q}then\frac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }}\) is equal to:1). \(\frac{{{P^2} - {q^2}}}{{{p^2} + {q^2}}}\)2). \(\frac{{{P^2} - {q^2}}}{{2pq}}\)3). \(\frac{{2pq}}{{{p^2} - {q^2}}}\)4). \(\frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}}\) |
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Answer» Given, $(\FRAC{{\SIN \theta }}{{\COS \theta }} = \frac{p}{q})$ $(\cos \theta = \sin \theta\times \frac{q}{p})$ ⇒ $(? = \frac{{p\sin \theta- q\cos \theta }}{{p\sin \theta + q\cos \theta }})$ ⇒ $(? = \frac{{\left( {\frac{p}{q}\sin \theta - \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}}{{\left( {\frac{p}{q}\sin \theta + \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}})$ ⇒ $(? = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}})$ |
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