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If θ lies in the second quadrant, then show that 1-sinθ1+sinθ+1+sinθ1-sinθ = −2sec θ

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प्रश्न

If θ lies in the second quadrant, then show that `sqrt((1 - sin theta)/(1 + sin theta)) + sqrt((1 + sin theta)/(1 - sin theta))` = −2sec θ

योग
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उत्तर

We have

`sqrt((1 - sin theta)/(1 + sin theta)) + sqrt((1 + sin theta)/(1 - sin theta)) = (1 - sin theta)/sqrt(1 - sin^2theta) + (1 + sin theta)/sqrt(1 - sin^2theta)`

= `2/sqrt(cos^2theta)`

= `2/|cos theta|`  .....(Since `sqrt(alpha^2)` = |α| for every real number α)

Given that θ lies in the second quadrant

So |cos θ| = – cos θ .....(Since cos θ < 0).

Hence, the required value of the expression is `2/(-costheta) = - 2 sectheta`

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अध्याय 3: Trigonometric Functions - Solved Examples [पृष्ठ ४१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 3 Trigonometric Functions
Solved Examples | Q 4 | पृष्ठ ४१

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