Answer:
±17/13
Step-by-step explanation:
[tex]sec^2x-tan^2 x=1\\sec^{2} x-(\frac{5}{12} )^2=1\\sec^{2} x=1+\frac{25}{144} =\frac{169}{144}\\sec ~x=\pm\frac{13}{12} \\cos~x=\pm\frac{12}{13}\\x ~lies~in ~first~or~3rd~quadrant~as~tan~x>0\\if~ x~lies~ in~ 1st~ ,~then~ both sin~and~cos~are~positive.\\sin ~x=\sqrt{1-cos^2x} =\sqrt{1-(\frac{12}{13})^2} =\frac{5}{13}\\sin~x+cos~x=\frac{12}{13}+\frac{5}{13}=\frac{17}{13}\\if ~x~lies~in~3rd~quadrant~then~both~are~negative.\\sin~x+cos~x=-\frac{17}{13}[/tex]
