हिंदी

If m = (cos θ – sin θ) and n = (cos θ + sin θ), show that sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 – tan^2θ).

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प्रश्न

If `m = (cos θ - sin θ)` and `n = (cos θ + sin θ)`, show that `sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 - tan^2θ)`.

योग
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उत्तर

LBS = `sqrt(m/n) + sqrt(n/m)`

= `sqrt(m)/sqrt(n) + sqrt(n)/sqrt(m)`

= `(m+n)/sqrt(mn)`

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

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

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

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

= `2/sqrt(1- tan^2 theta)`

= RHS

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अध्याय 13: Trigonometric identities - EXERCISE 13В [पृष्ठ ६२९]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 13 Trigonometric identities
EXERCISE 13В | Q 10. | पृष्ठ ६२९

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