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Question Number 62577 by Sr@2004 last updated on 23/Jun/19

Commented by Sr@2004 last updated on 23/Jun/19

$$pleasesolve12$$

Answered by som(math1967) last updated on 23/Jun/19

$${(cscx-csczcscy)}^2={cot}^2{ycot}^{2}z★$$$${csc}^{2}x+{csc}^2{zcsc}^{2}y-2cscxcscycscz=({csc}^{2}y-1){cot}^{2}z$$$${csc}^{2}x+{csc}^{2}y(1+{cot}^{2}z)-2cscxcscycscz$$$$={cot}^2{ycsc}^{2}z-{cot}^{2}z$$$${csc}^{2}x+{csc}^{2}y-2cscxcscycscz=-{cot}^{2}z$$$${csc}^{2}x+{csc}^2{xcot}^{2}z+{csc}^{2}y-2cscxcscycscz$$$$=-{cot}^{2}z+{csc}^2{xcot}^{2}z★★$$$${csc}^{2}x(1+{cot}^{2}z)+{csc}^{2}y-2cscycsczcscx$$$$={cot}^{2}z({csc}^{2}x-1)$$$${csc}^2{xcsc}^{2}z+{csc}^{2}y-2cscycscxcscz={cot}^2{zcot}^{2}x$$$${(cscy-cscxcscz)}^2={cot}^2{zcot}^{2}x$$$$(cscy-cscxcscz)=\pm cotzcotx$$$$\therefore cosecy=cosecxcosecz\pm cotzcotx$$$$★csccosec$$$$★★add{cosec}^2{xcot}^{2}zbothside$$$$$$

Commented by Sr@2004 last updated on 24/Jun/19

$$icannotunderstand$$$$$$$$$$$$$$

Commented by som(math1967) last updated on 24/Jun/19

$$whichline?$$

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