If sinѲ = 12/13, find the value of (sec Ѳ + tan Ѳ) / ( sec Ѳ – tan Ѳ)
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sinФ=12/13=perpendicular/hypotenuse
then by Pythagoras's theorem,perpendicular²+base²=hypotaneous²
here, perpendicular=12, hypotenuse=13
∴, base=√hypotenuse²-perpendicular²=√169-144=√25=5
∴,cosФ=base/hypotenuse=5/13
∴, secФ=1/cosФ=1/(5/13)=13/5 and tanФ=sinФ/cosФ=(12/13)/(5/13)=12/5
∴, (secФ+tanФ)/(secФ-tanФ)=(13/5+12/5)/(13/5-12/5)=[(13+12)/5]/[(13-12)/5] = (25/5)/(1/5)=25
then by Pythagoras's theorem,perpendicular²+base²=hypotaneous²
here, perpendicular=12, hypotenuse=13
∴, base=√hypotenuse²-perpendicular²=√169-144=√25=5
∴,cosФ=base/hypotenuse=5/13
∴, secФ=1/cosФ=1/(5/13)=13/5 and tanФ=sinФ/cosФ=(12/13)/(5/13)=12/5
∴, (secФ+tanФ)/(secФ-tanФ)=(13/5+12/5)/(13/5-12/5)=[(13+12)/5]/[(13-12)/5] = (25/5)/(1/5)=25
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