Question

`"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")` हे सिद्ध करा.

Answer 1

उजवी बाजू = `(sec^2"A")/("cosec"^2"A")`

= `(1 + tan^2"A")/(1 + cot^2"A")`    .....`[(because 1 + tan^2"A" = sec^2"A"),(1 + cot^2"A" = "cosec"^2A")]`

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

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

= `(1/(cos^2"A"))/(1/(sin^2"A"))`     .......[∵ sin²A + cos²A = 1]

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

= tan²A

= tan A . tan A

= `"tan A"/"cot A"`

= डावी बाजू 

∴ `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`