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13, Prove thar, JICH % \[_\Mse 2cosecdsecO+1

Answer»

√(secA-1/secA+1)+√(secA+1/secA-1)=√{(secA-1)(secA-1)/(secA+1)(secA-1)}+√{(secA+1)(secA+1)/(secA-1)(secA+1)}=√{(secA-1)²/(sec²A-1)}+√{(secA+1)²/(sec²A-1)}=√(secA-1)²/tan²A+√(secA+1)²/tan²A=(secA-1)/tanA+(secA+1)/tanA=secA/tanA-1/tanA+secA/tanA+1/tanA=(1/cosA)/(sinA/cosA)-cotA+(1/cosA)/(sinA/cosA)+cotA=1/sinA+1/sinA=2/sinA=2cosecA (Proved)Please like the solution 👍 ✔️👍



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