Advertisements
Advertisements
प्रश्न
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Advertisements
उत्तर
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))`
= `sqrt((1 + sinq)/(1 - sinq) . (1+ sinq)/(1 + sinq)) + sqrt((1 - sinq)/(1 + sinq) . (1 - sinq)/(1 - sinq))`
= `sqrt((1 + sinq)^2/(1 - sin^2q)` + `sqrt((1 - sinq)^2/(1 - sin^2q))` = `sqrt((1 + sinq)^2/cos^2q)` + `sqrt((1 - sinq)^2/cos^2q)`
= `(1 + sinq)/cosq + (1 - sinq)/cosq = (1 + sinq + 1 - sinq)/cosq` = `2/cosq`
= 2 secq
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
If `sec theta + tan theta = x," find the value of " sec theta`
\[\frac{x^2 - 1}{2x}\] is equal to
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
If 1 + sin2α = 3 sinα cosα, then values of cot α are ______.
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
