Question

यदि tanθ = 2, तो अन्य त्रिकोणमितीय अनुपातों के मान ज्ञात कीजिए। 

Answer 1

 

मानो, ΔABC में,

∠ABC = 90° और ∠ACB = θ

tanθ = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

tanθ = 2

∴ ${\displaystyle \frac{\text{AB}}{\text{BC}}=\frac{2}{1}}$

∴ AB = 2k और BC = k

समकोण ΔABC में,

पायथागोरस के प्रमेय से,

${\displaystyle {\text{AC}}^2={\text{AB}}^2+{\text{BC}}^2}$

∴ ${\displaystyle {\text{AC}}^2={\left(2k\right)}^2+k^2}$

∴ ${\displaystyle {\text{AC}}^2=4k^2+k^2}$

∴ ${\displaystyle {\text{AC}}^2=5k^2}$

∴ AC = ${\displaystyle \sqrt{5}}$.k .................(दोनों पक्षों का वर्गमूल लेने पर)

sinθ = ${\displaystyle \frac{\text{AB}}{\text{AC}}=\frac{2k}{\sqrt{5}.k}=\frac{2}{\sqrt{5}}}$

cosθ = ${\displaystyle \frac{\text{BC}}{\text{AC}}=\frac{k}{\sqrt{5}.k}=\frac{1}{\sqrt{5}}}$

cosecθ = ${\displaystyle \frac{\text{AC}}{\text{AB}}=\frac{\sqrt{5}.k}{2k}=\frac{\sqrt{5}}{2}}$

secθ = ${\displaystyle \frac{\text{AC}}{\text{BC}}=\frac{\sqrt{5}.k}{k}=\frac{\sqrt{5}}{1}}$

cotθ = ${\displaystyle \frac{\text{BC}}{\text{AB}}=\frac{k}{2k}=\frac{1}{2}}$

sinθ = ${\displaystyle \frac{2}{\sqrt{5}}}$, cosθ = ${\displaystyle \frac{1}{\sqrt{5}}}$, cosecθ = ${\displaystyle \frac{\sqrt{5}}{2}}$, secθ = ${\displaystyle \frac{\sqrt{5}}{1}}$, cotθ = ${\displaystyle \frac{1}{2}}$