Question

`(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")` हे सिद्ध करा. 

Answer 1

डावी बाजू = `(1 + sec "A")/"sec A"`

= `1/"sec A" + "sec A"/"sec A"`

= cos A + 1

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

= `(1 - cos^2"A")/(1 - cos"A")`

= `(sin^2"A")/(1 - cos"A")`     .......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2"A")]`

= उजवी बाजू

∴ `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`