मराठी

Prove that 1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ

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प्रश्न

Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`

बेरीज
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उत्तर

L.H.S = `(1 + sec theta - tan theta)/(1 + sec theta + tan theta)`

= `(1 + 1//cos theta - sin theta//cos theta)/(1 + 1//cos theta + sin theta//cos theta)`  ...`[∵ sec theta = 1/cos theta and tan theta = sin theta/cos theta]`

= `(cos theta + 1 - sin theta)/(cos theta + 1 + sin theta)`

= `((cos theta + 1) - sin theta)/((cos theta + 1) + sin theta)`

= `(2 cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2 cos^2  theta/2 + 2 sin  theta/2 * cos  theta/2)`  ...`[∵ 1 + cos theta = 2 cos^2  theta/2 and sin theta = 2sin  theta/2 cos  theta/2]`

= `(2cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2cos^2  theta/2 + 2sin  theta/2 * cos  theta/2)`

= `(2cos  theta/2 (cos  theta/2 - sin  theta/2))/(2cos  theta/2(cos  theta/2 + sin  theta/2))`

= `(cos  theta/2 - sin  theta/2)/(cos  theta/2 + sin  theta/2) xx ((cos  theta/2 - sin  theta/2))/((cos  theta/2 - sin  theta/2))`  ...[By rationalisation]

= `(cos  theta/2 - sin  theta/2)^2/((cos^2  theta/2 - sin^2  theta/2))`  ...[∵ (a – b)2 = a2 + b2 – 2ab and (a – b)(a + b) = (a2 – b2)]

= `((cos^2  theta/2 + sin^2  theta/2) - (2 sin  theta/2 * cos  theta/2))/cos theta`   ...`[∵ cos^2  theta/2 - sin^2  theta/2 = cos theta]`

= `(1 - sin theta)/cos theta`   ...`[∵ sin^2  theta/2 + cos^2  theta/2 = 1]`

= R.H.S

Hence proved.

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पाठ 8: Introduction To Trigonometry and Its Applications - Exercise 8.4 [पृष्ठ ९९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 10
पाठ 8 Introduction To Trigonometry and Its Applications
Exercise 8.4 | Q 12 | पृष्ठ ९९

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