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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B - Geometry Mathematics 2

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प्रश्न

Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B

बेरीज

उत्तर

L.H.S = (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

= R.H.S

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

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