Advertisements
Advertisements
प्रश्न
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Advertisements
उत्तर
L.H.S. = (1 + cot A – cosec A)(1 + tan A + sec A)
= `(1 + cosA/sinA - 1/sinA)(1 + sinA/cosA + 1/cosA)`
= `((sinA + cosA - 1)/sinA)((cosA + sinA + 1)/cosA)`
= `((sinA + cosA - 1)(sinA + cosA + 1))/(sinAcosA)`
= `((sinA + cosA)^2 - (1)^2)/(sinAcosA)`
= `(sin^2A + cos^2A + 2sinAcosA - 1)/(sinAcosA)`
= `(1 + 2sinAcosA - 1)/(sinAcosA)`
= `(2sinAcosA)/(sinAcosA)`
= 2 = R.H.S.
संबंधित प्रश्न
Prove the following identities:
`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Evaluate sin25° cos65° + cos25° sin65°
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
If sec θ + tan θ = x, then sec θ =
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
