Question

यदि cosec θ + cot θ = p है, तो सिद्ध कीजिए कि cos θ = ${\displaystyle \frac{p^2-1}{p^2+1}}$ है। 

Answer 1

प्रश्न के अनुसार,

cosec θ + cot θ = p

चूँकि, cosec θ = ${\displaystyle \frac{1}{\sin \theta }}$ और cot θ = ${\displaystyle \frac{\cos \theta }{\sin \theta }}$

${\displaystyle \frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }}$ = p

${\displaystyle \frac{1+\cos \theta }{\sin \theta }}$ = p

L.H.S और R.H.S पर वर्ग बनाना,

${\displaystyle {\left(\frac{1+\cos \theta }{\sin \theta }\right)}^2}$ = p²

${\displaystyle \frac{1+{\cos }^2\theta +2\cos \theta }{{\sin }^2\theta }}$ = p²

घटक और लाभांश नियम लागू करना,

${\displaystyle \frac{(1+{\cos }^2\theta +2\cos \theta )-{\sin }^2\theta }{(1+{\cos }^2\theta +2\cos \theta )+{\sin }^2\theta }=\frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

= ${\displaystyle \frac{(1-{\sin }^2\theta )+{\cos }^2\theta +2\cos \theta }{{\sin }^2\theta +{\cos }^2\theta +1+2\cos \theta }=\frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

चूँकि, 1 – sin²θ = cos²θ and sin²θ + cos²θ = 1

${\displaystyle \frac{{\cos }^2\theta +{\cos }^2\theta +2\cos \theta }{1+1+2\cos \theta }=\frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

${\displaystyle \frac{2{\cos }^2\theta +2\cos \theta }{2+2\cos \theta }=\frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

${\displaystyle \frac{2\cos \theta (\cos \theta +1)}{2(\cos \theta +1)}=\frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

cos θ = ${\displaystyle \frac{{\text{p}}^2-1}{{\text{p}}^2+1}}$

अतः सिद्ध हुआ।