Advertisements
Advertisements
प्रश्न
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
Advertisements
उत्तर
L.H.S = `sqrt(sec^2 theta + "cosec"^2 theta)`
= `sqrt(1/cos^2 theta + 1/(sin^2 theta))` ...`[∵ sec^2 theta = 1/(cos^2 theta) "and" "cosec"^2 theta = 1/(sin^2 theta)]`
= `sqrt((sin^2 theta + cos^2 theta)/(sin^2 theta * cos^2 theta))`
= `sqrt(1/(sin^2 theta * cos^2 theta))` ...[∵ sin2θ + cos2θ = 1]
= `1/(sin theta * cos theta)`
= `(sin^2 theta + cos^2 theta)/(sin theta * cos theta)` ...[∵ 1 = sin2θ + cos2θ]
= `(sin^2 theta)/(sin theta * cos theta) + (cos^2 theta)/(sin theta * cos theta)`
= `sintheta/costheta + cos theta/sintheta` ...`[∵ tan theta = sin theta/cos theta "and" cot theta = costheta/sin theta]`
= tan θ + cot θ
= R.H.S
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
If `sin theta = x , " write the value of cot "theta .`
What is the value of (1 − cos2 θ) cosec2 θ?
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
sec4 A − sec2 A is equal to
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Choose the correct alternative:
sec2θ – tan2θ =?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
If 5 tan β = 4, then `(5 sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.
