Question

underoot of Sec square+Cosec square=tan+ Cot

Answer 1

Answer:

$\sqrt{sec {}^{2} + cosec {}^{2} } = tan \: + cot \\ lhs = \sqrt{ \frac{1}{cos {}^{2} } + \frac{1}{sin { }^{2} } } \\ = \sqrt{ \frac{sin {}^{2} + cos {}^{2} }{sin {}^{2} cos {}^{2} } } \\ = \sqrt{( \frac{sin + cos}{sin \: cos} ) {}^{2} } \\ = \frac{sin + cos}{sin \: cos} \\ = \frac{sin}{sin \: cos } + \frac{cos}{sin \: cos} \\ = \frac{sin}{cos} + \frac{cos}{sine} \\ = tan \: + cot \\ = rhs$

Hence proved

Answer 2

Step-by-step explanation:

Given :

√sec²∅ + cosec²∅

=> √(1 + tan²∅) + (1 + cot²∅)

=> √1 + tan²∅ + 1 + cot²∅

=> √tan²∅ + cot²∅ + 2

=> √tan²∅ + cot²∅ + 2 * 1

=> √tan²∅ + cot²∅ + 2 * tan∅ * cot∅

=> √(tan∅ + cot∅)²

=> tan∅ + cot∅

Hope it helps!