यदि ` see theta + tan theta = p ` तो सिद्ध कीजिए कि : ` (p^(2) - 1)/(p^(2)+ 1) = sin theta `

Explanation

L.H.S = ` (p^(2) - 1)/(p^(2)+ 1) = ((sec theta + tan theta )^(2)-1)/((sec theta + tan theta )^(2)+1)= (sec^(2) theta + tan^(2)theta + 2 sec theta tan theta - 1)/(sec^(2) theta + tan^(2) theta + 2 sec theta tan theta + 1)`
`((sec^(2) theta - 1 ) + tan^(2) theta + 2 sec theta tan theta )/(sec^(2) theta + (tan^(2) theta + 1) + 2 sec theta tan theta ) = (tan^(2) theta + tan^(2) theta + 2 sec theta tan theta )/(sec^(2) theta + sec^(2) theta + 2 sectheta tan theta ) `
` (2 tan^(2) theta + 2 sec theta tan theta )/(2 sec ^(2) theta + 2 sec theta tan theta) = (2 tan theta (tan theta + sec theta))/(2 sec theta (sec theta + tan theta)) = (tan theta )/(sec theta) = (sin theta )/(costheta ) xx(cos theta )/(1)`
` sin theta = R.H.S`
वैकल्पिक विधि : यहाँ ` sec theta + tan theta = p `
` rArr (1)/(cos theta ) + (sin theta )/(cos theta ) = (p)/(1) rArr (1 + sin theta )/(cos theta ) = (p)/(1) `
यहाँ दोनों पक्षों का वर्ग करने पर
` ((1 + sin theta)^(2))/(cos^(2) theta ) = (p^(2))/(1) rArr ((1+ sin theta)^(2))/(1 - sin ^(2) theta) = (p^(2))/(1) (sin ^(2) theta cos^(2) theta = 1 से )`
`rArr ((1 + sin theta )^(2))/((1 + sin theta ) (1 - sin theta )) = (p^(2))/(1) rArr (1 + sin theta)/(1 - sin theta ) = (p^(2))/(1)`
योगान्तर अनुपात के नियम से
` (2)/(2 sin theta ) = (p^(2) + 1)/(p^(2)- 1)`
` rArr (1) /(sin theta) = (p^(2) + 1)/(p^(2) - 1) rArr sin theta = (p^(2) - 1)/(p^(2) + 1)`