Question

Answer fast as you can.​

Answer 1

$\LARGE\star\mathfrak{\underline{\underline{\red{ǫᴜᴇsᴛɪᴏɴ: }}}}$

$\sf{prove \: that \: \frac{1}{sin {}^{2} \theta } - \frac{1}{sin {}^{2} \phi } } = \frac{cos {}^{2} \theta - cos {}^{2} \phi }{sin {}^{2} \theta.sin {}^{2} \phi }$

$\LARGE\star\mathfrak{\underline{\underline{\red{answer: }}}}$

Taking LHS :-

$\sf{ \frac{1}{sin {}^{2} \theta } - \frac{1}{sin \phi} } \\ \\ \sf{ = \frac{sin {}^{2} \phi - sin {}^{2} \theta }{sin {}^{2} \theta.sin {}^{2} \phi } } \\ \\ \sf{identity :sin {}^{2} \theta = 1 - cos {}^{2} \theta} \\ \\ \sf{ \frac{(1 - cos {}^{2} \phi) - (1 - cos {}^{2} \theta )}{sin {}^{2} \theta.sin {}^{2} \phi } } \\ \\ \sf{ \frac{1 - cos { }^{2} \phi - 1 + cos {}^{2} \theta }{sin {}^{2} \theta.sin {}^{2} \phi} } \\ \\ \sf{ \frac{cos {}^{2} \theta - cos {}^{2} \phi}{sin {}^{2} \theta.sin {}^{2} \phi} } \\ \\ \sf{ʟʜs = ʀʜs} \\ \sf{hence \: proved}$