Advertisements
Advertisements
प्रश्न
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove the following:
`(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Advertisements
उत्तर
LHS = `(sec θ - tan θ)/(sec θ + tan θ )`
= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`
= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`
= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`
= 1 + 2 tan2θ − 2 sec θ. tan θ
= R.H.S.
Hence proved.
APPEARS IN
संबंधित प्रश्न
9 sec2 A − 9 tan2 A = ______.
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
9 sec2 A − 9 tan2 A is equal to
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
