Question

Prove that sin theta minus cos secant theta whole square + cos theta minus secant theta whole square is equals to COT square theta + tan squared theta minus one

Answer 1

$( {sin \theta - cosec \theta})^{2} + ( {cos \theta - sec \theta})^{2} \\ = {sin}^{2} \theta + {cosec}^{2} \theta - 2 + {cos}^{2} \theta + {sec}^{2} \theta - 2 \\ = 1 - 4 + {cosec}^{2} \theta + {sec}^{2} \theta \\ = - 3 + 1 + {cot}^{2} \theta + 1 + {tan}^{2} \theta \\ = {tan}^{2} \theta + {cot}^{2} \theta - 1$