Question

pls help its urgent​

Answer 1

Answer:

To prove:-

$\sqrt{ \dfrac{1 + {sin}^{2} \theta. {sec}^{2} \theta}{1 + {cos}^{2}\theta. {cosec}^{2}\theta} } = tan \theta$

Properties used:-

$sec\theta = \dfrac{1}{cos\theta} \\ \\ cosec\theta = \dfrac{1}{sin\theta} \\ \\ 1 + {tan}^{2} \theta = {sec}^{2} \theta \\ \\ 1 + {cot}^{2} \theta = {cosec}^{2} \theta$

LHS:-

$\sqrt{ \dfrac{1 + {sin}^{2} \theta. {sec}^{2} \theta}{1 + {cos}^{2}\theta. {cosec}^{2}\theta} } \\ \\ = \sqrt{ \dfrac{1 + {sin}^{2} \theta. \dfrac{1}{ {cos }^{2} \theta} }{1 + {cos}^{2}\theta. \dfrac{1}{ {sin}^{2}\theta} } } \\ \\ = \sqrt{ \dfrac{1 + {tan}^{2}\theta}{1 + {cot}^{2}\theta} } \\ \\ = \sqrt{ \frac{sec {}^{2}\theta}{ {cosec}^{2}\theta}} \\ \\ = \frac{sec \theta}{cosec \theta} \\ \\ = \dfrac{ \dfrac{1}{cos \theta } }{ \dfrac{1}{sin\theta }} \\ \\ = \frac{sin \theta}{cos\theta} \\ \\ = tan\theta$