Advertisements
Advertisements
प्रश्न
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Advertisements
उत्तर
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
`sqrt((1 + sin theta)/(1 - sin theta)) = sqrt(((1 + sin theta)(1 + sin theta))/((1 - sin theta)(1 + sin theta))`
= `sqrt((1 + sin theta)^2/(1 - sin^2 theta)`
= `sqrt((1 + sin theta)^2/(cos^2 theta)`
= `(1 + sin theta)/cos theta`
`sqrt(((1 - sin theta))/((1 + sin theta))) = sqrt(((1 - sin theta))/((1 - sin theta)) xx ((1 + sin theta))/((1 - sin theta))`
= `sqrt((1 - sin theta)^2/(1 - sin^2 theta)`
= `sqrt((1- sin theta)^2/(cos^2 theta)) = (1 - sin theta)/cos theta`
L.H.S. = `sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta)`
= `(1 + sin theta)/cos theta + (1 - sin theta)/cos theta`
= `(1 + sin theta + 1 - sin theta)/cos theta`
= `2/cos theta`
= 2 sec θ
L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
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)`.
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
