Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Advertisements
उत्तर
LHS = `(secA - 1)/(secA + 1)`
= `(1/cosA - 1)/(1/cosA + 1) = (1 - cosA)/(1 + cosA)`
= `(1 - cosA)/(1 + cosA) xx (1 + cosA)/(1 + cosA)`
= `(1-cos^2A)/(1 + cosA)^2`
= `sin^2A/(1 + cosA)^2` (∵ `1 - cos^2A = sin^2A`)
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
sin2θ + sin2(90 – θ) = ?
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.
