Advertisements
Advertisements
प्रश्न
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
Advertisements
उत्तर
L.H.S. = sec2θ + cosec2θ
= `1/(cos^2θ) + 1/(sin^2θ)`
= `(sin^2θ + cos^2θ)/(cos^2θ.sin^2θ)`
= `1/(cos^2θ.sin^2θ)` ...[∵ sin2θ + cos2θ = 1]
= `1/(cos^2θ) xx 1/(sin^2θ)`
= sec2θ × cosec2θ
= R.H.S.
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
`(1 + cot^2A)/(1 + tan^2A)` = ?
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
If 2sin2θ – cos2θ = 2, then find the value of θ.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
(1 + sin A)(1 – sin A) is equal to ______.
