If $tan x =\frac{m}{n}$ and 0° ≤ x ≤ 90°, then the value of (sinx + cos x) is :

$\frac{m+n}{\sqrt{m^2+n^2}}$

tanx = \(\frac{m }{n}\)

{ tanx = \(\frac{P }{B}\) }

By using pythagoras theorem ,

P² + B² = H²

m² + n² = H²

H = \(\sqrt {m² + n² }\)

Now,

sinx + cosx

= \(\frac{P }{H}\) + \(\frac{B }{H}\)

= $\frac{m}{\sqrt{m^2+n^2}}$ + $\frac{n}{\sqrt{m^2+n^2}}$

= $\frac{m+n}{\sqrt{m^2+n^2}}$