Question

If m=cosq-sinq and n=cosq+sinq,then show that√(m/n) + √(n/m) = 2/√(1-tan2q)

Answer 1

LHS = $\bold{\sqrt{\frac{m}{n}}}+\bold{\sqrt{\frac{n}{m}}}$
$=\bold{\sqrt{\frac{m}{n}}}+\bold{\sqrt{\frac{n}{m}}}\\\\=\bold{\frac{(m+n)}{\sqrt{mn}}}$

now, m = cosq - sinq , n = cosq + sinq
( m + n) = (cosq - sinq) + (cosq + sinq )
= 2cosq

mn = (cosq - sin)(cosq + sinq) = cos²q - sin²q [ as you know, a² -b² = (a -b)(a + b)]
= (cos²q - sin²q)/(sin²q + cos²q) [ as we know, sin²x + cos²x = 1 ]
= (1 - tank)/(1 + tan²q) [ after dividing with cos²q both numerator and denominator]

now,
LHS = (m + n)/√mn
= 2cosq/√{(1 - tan²q)/(1 + tan²q)}
= 2cosq/√{(1 - tan²q)/sec²q} [ sec²x = 1 + tan²x]
= 2cosq.secq/√(1 - tan²q)
= 2/√(1 - tan²q) = RHS