Advertisements
Advertisements
प्रश्न
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
उत्तर
Given cot θ + tan θ = x and sec θ – cos θ = y
x = cot θ + tan θ
x = `1/tan theta + tan theta`
= `(1 + tan^2 theta)/tan theta`
= `(sec^2 theta)/tan theta`
= `(1/cos^2theta)/(sin theta/costheta`
= `1/(cos theta sin theta)`
y = sec θ – cos θ
= `1/cos theta - cos theta`
= `(1 - cos^2 theta)/cos theta`
y = `(sin^2 theta)/costheta`
= `[1/(cos^2thetasin^2theta) xx (sin^2theta)/costheta]^(2/3) - [1/(cos theta sin theta) xx (sin^4 theta)/(cos^2 theta)]^(2/3)`
= `[1/(cos^3theta)]^(2/3) - [(sin^3 theta)/(cos^3 theta)]^(2/3)`
= `[1/(cos^2 theta)] - [(sin^2 theta)/(cos^2 theta)]`
= `[(1 - sin^2 theta)/(cos^2 theta)]`
= `[(cos^2 theta)/(cos^2 theta)]`
= 1
L.H.S = R.H.S
⇒ `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
Evaluate:
`(tan 65°)/(cot 25°)`
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
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 ______.
