Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Advertisements
उत्तर
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2`
= `(sinθ/cosθ + 1/cosθ)^2 + (sinθ/cosθ - 1/cosθ)^2`
= `((sinθ + 1)/cosθ)^2 + ((sinθ - 1)/cosθ)^2`
= `(sinθ + 1)^2/(cos^2θ) + (sinθ - 1)^2/cos^2θ`
= `((sinθ + 1)^2 + (sinθ - 1)^2)/cos^2A`
= `(sin^2θ + 1 + 2sinθ + sin^2θ + 1 - 2sinθ)/(1 - sin^2θ)`
= `(2(1 + sin^2θ))/(1 - sin^2θ)`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
If 2sin2θ – cos2θ = 2, then find the value of θ.
Show that tan4θ + tan2θ = sec4θ – sec2θ.
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
