(i) यदि `cos theta = q/sqrt(p^(2) + q^(2)), `तो सिद्ध करें कि `(sqrt(p^(2)+q^(2))/p + q/p)^(2) = (sqrt(p^(2)+q^(2))+q)/(sqrt(p^(2)+q^(2))-q)`

Explanation

R.H.S. ` = (1 + q/(sqrt(p^(2) + q^(2))))/(1 - q/(sqrt(p^(2)+q^(2))) ) = (1 + cos theta)/(1- cos theta) [ "अंश और हर को " sqrt(p^(2)+q^(2))" से भाग देने पर "]`
` = ( 1+ cos theta)/(1 - cos theta) xx ( 1 + cos theta)/(1 + cos theta) = ((1 - cos theta)^(2))/(1- cos^(2) theta) = ((1 + cos theta)^(2))/( sin^(2) theta) `
` = ((1+ cos theta)/(sin theta))^(2) =(1/(sin theta) +(cos theta)/(sin theta))^(2) = (cosec theta + cot theta)^(2)` ....(i)
` :' cos theta = q/sqrt(p^(2) + q^(2)) = b/h`
हम समकोण त्रिभुज ABC खींचते है , जिसमें
`angleACB = theta, BC = q, AC = sqrt(p^(2) + q^(2))`
अब `AB = sqrt(AC^(2) - BC^(2)) = sqrt((sqrt(p^(2) + q^(2)))^(2) - q^(2)) = sqrt(p^(2) + q^(2) - q^(2)) = p `
` :. cosec theta = h/p = (AC)/(AB) = sqrt(p^(2) + q^(2))/p " और " cot theta = b/p = q/p`
समीo (i) R.H.S. = ` (sqrt(p^(2)+q^(2))/p + q/p)` = L.H.S.]