($ sinpx + cospx)^2 \;=\; sin^2px+cos^2px+2sinpx \cdot cospx \\ \; \\ = 1+ sin2px \left[ \begin{array}{c} \because sin^2\theta + cos^2\theta=1 \& \\ 2sin\theta cos\theta \;=\; sin2\theta \end{array} \right] \\ \; \\ \; \\ \therefore sinpx+cospx \; = \; (1+ sin2px)^{1/2} \\ \; \\ \; \\ \therefore y\;=\; (1+ sin2px)^{1/2} \\ \; \\ \; \\ \; \\ Diff \; y \; w.r.t. \; x, \\ \; \\ \; \\ \therefore y_1 \;=\; pcospx\;-\;psinpx \;=\; p[cospx\;+\;sinpx] \; \; \ldots (i) \\ \; \\ Now, \; (cospx \;-\; sinpx)^2 \;=\; cos^2px + sin^2px -2sinpxcospx \; \; \ldots (ii) \\ \; \\ \; \\ = 1-sin2px \;=\; 1+ (-1)^1 sin2px \\ \; \\ \therefore y_1 \;=\; p^1 [1+(-1)^1 sin2px] \\ \; \\ \; \\ \; \\ Diff. \; equation \; (i) \; w.r.t. \; x, \\ \; \\ \; \\ y_2 \;=\; -p \cdot p sinpx \;+\; p \cdot pcospx \;=\; -p^2 sinpx \;+\; p^2 cospx \;\;\;\;\; \therefore y_2\;=\; p^2[-sinpx+cospx] \;\;\;\ldots (ii) \\ \; \\ \; \\ Now, (-cospx \;+\; sinpx)^2 \;=\; \Big[ (-1)^2 (cospx \;+\; sinpx)^2 \Big] \;=\; (1+sin2px) \\ \; \\ \; \\ \therefore (-sinpx+cospx) \;=\; \Big[ 1+(-1)^2 sin2px \Big]^{1/2} \\ \; \\ \; \\ \therefore y_2 \;=\; p^2 \Big[ 1+(-1)^2 sin2px \Big]^{1/2} \\ \; \\ \; \\ \; \\ Diff \; again \; equation \; (ii) \; w.r.t. \; x \\ \; \\ \; \\ \therefore y_2\;=\; p^2[-cospx+sinpx] \\ \; \\ \; \\ Now, (-cospx+sinpx)^2 \;= \; cos^2px + sin^2px -2sinpxcospx \; = \; 1-sin2px \\ \; \\ \; \\ \; \\ \therefore -cospx+sinpx \;=\; [1-sin2px]^{1/2} \;=\; [1+(-1)^3 sin2px]^{1/2} \\ \; \\ \; \\ \therefore y_3 \;=\; p^3 [1+(-1)^3sin2px]^{1/2} \\ \; \\ \;\\ y_n \;=\; p^n [1+(-1)^n sin2px ]^{1/2} \; \; \; \; Hence \; Proved. $