Advertisements
Advertisements
प्रश्न
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
Advertisements
उत्तर
LHS = `sinA/sin(90^circ - A) + cosA/cos(90^circ - A)`
⇒ `cos(90^circ - A)/sin(90^circ - A) + sin(90^circ - A)/cos(90^circ - A)`
⇒ `(cos^2(90^circ - A) + sin^2(90^circ - A))/(sin(90^circ - A) . cos(90^circ - A)) = 1/(sin(90^circ - A) . cos(90^circ - A))`
⇒ `sec(90^circ - A) cosec(90^circ - A)`
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
