Question

`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1 हे सिद्ध करा.

Answer 1

`"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")`

= `((cos "A")/(sin "A"))/(1 - (sin "A")/(cos "A")) + ((sin "A")/(cos "A"))/(1 - (cos "A")/(sin "A"))`

= `((cos "A")/(sin "A"))/((cos "A" -  sin "A")/(cos "A")) + ((sin "A")/(cos "A"))/((sin "A" -  cos "A")/(sin "A"))`

= `"cos A"/"sin A" xx "cos A"/(cos "A" - sin "A") + "sin A"/"cos A" xx "sin A"/(sin "A" - cos "A")`

= `(cos^2"A")/(sin "A"(cos "A" - sin "A")) + (sin^2"A")/(cos"A"(sin"A" - cos"A"))`

= `1/(sin "A" - cos "A") ((-cos^3"A" + sin^3"A")/(sin"A" cos"A"))`

= `1/(sin"A" - cos"A")((sin^3"A" - cos^3"A")/(sin"A" cos"A"))`

= `1/(sin"A" - cos"A")xx ((sin"A" - cos"A")(sin^2"A" + sin"A" cos"A" + cos^2"A"))/(sin"A" cos"A")`  ......[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= `(sin^2"A" +sin"A" cos"A" + cos^2"A")/(sin"A" cos"A"`  ......(i)

= `(1 + sin"A" cos"A")/(sin"A" cos"A")`   .....[∵ sin²A + cos²A = 1]

= `1/(sin"A" cos"A") + (sin"A" cos"A")/(sin"A" cos"A")`

= cosec A sec A + 1  .....(ii)

`"cot A"/(1 - tan "A") + "tan A"/(1 - cot "A")`

= `(sin^2"A" + sin"A" cos"A" + cos^2"A")/(sin"A" cos"A")`     ......[(i) वरून]

= `(sin^2"A")/(sin"A" cos"A") + "sin A cos A"/"sin A cos A" + (cos^2"A")/"sin A cos A"`

= `"sin A"/"cos A" + 1 + "cos A"/"sin A"`

= tan A + 1 + cot A    ......(iii)

(ii) आणि (iii) वरून,

`"cot  A"/(1 - tan "A") + "tan A"/(1 - cot "A")` = 1 + tan A + cot A = sec A . cosec A + 1