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Prove that ( 1 + Sin θ)/(1 - Sin θ) = 1 + 2 Tan θ/Cos θ + 2 Tan^2 θ - Mathematics

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प्रश्न

Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .

योग

उत्तर

RHS = `1 + 2 tan θ/cos θ + 2 tan^2 θ`

= `1 + 2 sin θ/cos^2θ + 2 sin^2 θ/cos^2 θ`

= `(cos^2 θ + 2sin θ + 2 sin^2 θ)/(cos^2θ)`

= `(1 - sin^2θ + 2 sin θ + 2 sin^2θ )/(1 - sin^2θ)`

= `(1 + sin^2θ + 2 sin θ)/(1 - sin^2θ)`

= `(1 + sin θ)^2/( 1 + sin θ)(1 - sin θ)`

= `(1 + sin θ)/(1 - sin θ)`

= LHS
Hence proved.

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अध्याय 18: Trigonometry - Exercise 2

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आईसीएसई Mathematics [English] Class 10
अध्याय 18 Trigonometry
Exercise 2 | Q 16
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