KeerthanaPosted on यदि θ hetaθ न्यून कोण हो और cosθ=1517cos heta=frac{15}{17}cosθ=1715, तो cot(90∘−θ)cot left(90^{circ}- heta ight)cot(90∘−θ) का मूल्य है a2815frac{2 sqrt{8}}{15}1528 b217frac{sqrt{2}}{17}172 c815frac{8}{15}158 d8217frac{8 sqrt{2}}{17}1782 Answer : Option AExplanation : cosθ=1517⇒secθ=1cosθ=1715cos heta=frac{15}{17} Rightarrow sec heta=frac{1}{cos heta}=frac{17}{15}cosθ=1715⇒secθ=cosθ1=1517 ∴cot(90∘−θ)=tanθ=sec2θ−1 herefore cot left(90^{circ}- heta ight)= an heta=sqrt{sec ^{2} heta-1}∴cot(90∘−θ)=tanθ=sec2θ−1 =(1715)2−1=289225−1=sqrt{left(frac{17}{15} ight)^{2}-1}=sqrt{frac{289}{225}-1}=(1517)2−1=225289−1 =289−225225=64225=815=sqrt{frac{289-225}{225}}=sqrt{frac{64}{225}}=frac{8}{15}=225289−225=22564=158 Alternative: यह देखते हुए कि cosθ=1517=bhcos heta=frac{15}{17}=frac{b}{h}cosθ=1715=hb ऊँचाई =( कर्ण)2−(आधार)2=sqrt{( ext { कर्ण})^{2}-( ext {आधार})^{2}}=( कर्ण)2−(आधार)2 =(17)2−(15)2=64=8=sqrt{(17)^{2}-(15)^{2}}=sqrt{64}=8=(17)2−(15)2=64=8 cot(90∘−θ)=tanθ= Height Base cot left(90^{circ}- heta ight)= an heta=frac{ ext { Height }}{ ext { Base }}cot(90∘−θ)=tanθ= Base Height =815=frac{8}{15}=158 Rate This:NaN / 5 - 1 votesAdd comment