Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
Advertisements
उत्तर
L.H.S. = `(secA - tanA)/(secA + tanA)`
= `(secA - tanA)/(secA + tanA) xx (secA - tanA)/(secA - tanA)`
= `(secA - tanA)^2/(sec^2A - tan^2A)`
= `(sec^2A + tan^2A - 2secAtanA)/1`
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 A = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
