Advertisements
Advertisements
प्रश्न
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Advertisements
उत्तर
L.H.S. = `(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2`
= `(sinA/cosA + 1/cosA)^2 + (sinA/cosA - 1/cosA)^2`
= `((sinA + 1)/cosA)^2 + ((sinA - 1)/cosA)^2`
= `(sinA + 1)^2/(cos^2A) + (sinA - 1)^2/(cos^2A)`
= `((sinA + 1)^2 + (sinA - 1)^2)/(cos^2A)`
= `(sin^2A + 1 + 2sinA + sin^2A + 1 - 2sinA)/cos^2A`
= `(2sin^2A + 2)/(cos^2A)`
= `(2(1 + sin^2A))/(1 - sin^2A)`
= `2((1 + sin^2A)/(1 - sin^2A))` = R.H.S.
संबंधित प्रश्न
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
