Advertisements
Advertisements
प्रश्न
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
उत्तर
LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`
= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`
= `(sec θ + 1)/(sec θ - 1)`
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
