Question

if m=(cos theta-sin theta) and n=(cos theta+sin theta) then show that √m/n+√n/m=2/√1-tan²theta

Answer 1

m = (cos∅ - sin∅)
n = (cos∅ + sin∅)

$LHS = \sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}} \\ \\ = \frac{ \sqrt{m} }{ \sqrt{n} } + \frac{ \sqrt{n} }{ \sqrt{m} } \\ \\ = \frac{m + n}{ \sqrt{mn} }$
now, put m =( cos∅ - sin∅) and n = (cos∅ + sin∅)
= {(cos∅ - sin∅)+(cos∅ + sin∅)}/√{cos∅-sin∅)(cos∅+sin∅)}
= 2cos∅/√{cos²∅ - sin²∅}
= 2/√{cos²∅/cos²∅ - sin²∅/cos²∅}
= 2/√{1 - tan²∅} = RHS

Answer 2

hope it helps u and mark me good answer