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Prove that sinθsecθ+1+sinθsecθ-1 = 2 cot θ - Geometry Mathematics 2

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प्रश्न

Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ

योग
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उत्तर

L.H.S = `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` 

= `sintheta/(1/costheta + 1) + sintheta/(1/costheta - 1`

= `sintheta/((1 + costheta)/costheta) + sintheta/((1 - costheta)/(costheta))`

= `(sintheta costheta)/(1 + costheta) + (sintheta costheta)/(1 - costheta)`

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

= `sintheta costheta [(1 - costheta + 1 + costheta)/((1 + costheta)(1 - costheta))]`

= `sintheta costheta (2/(1 - cos^2theta))`   ......[∵ (a + b)(a – b) = a2 – b2]

= `sintheta costheta xx 2/(sin^2theta)`   .....`[(because sin^2theta + cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`

= `2 xx (costheta)/(sintheta)`

= 2cot θ

= R.H.S

∴ `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ

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अध्याय 6: Trigonometry - Q.3 (B)

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