Question

(1 – cos²A) . sec²B + tan²B (1 – sin²A) = sin²A + tan²B हे सिद्ध करा.

Answer 1

डावी बाजू = (1 – cos²A) . sec²B + tan²B(1 – sin²A)

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

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

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

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

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

= sin²A + tan²B

= उजवी बाजू

∴ (1 – cos²A) . sec²B + tan²B (1 – sin²A) = sin²A + tan²B