Advertisements
Advertisements
प्रश्न
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Advertisements
उत्तर
L.H.S. = `(cosecA - sinA)(secA - cosA)`
= `(1/sinA - sinA)(1/cosA - cosA)`
= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA)`
= `(cos^2A/sinA)(sin^2A/cosA)`
= sin A cos A
R.H.S. = `1/(tanA + cotA)`
= `1/(sinA/cosA + cosA/sinA)`
= `1/((sin^2A + cos^2A)/(sinAcosA))`
= `(sinAcosA)/(sin^2A + cos^2A)`
= `(sinAcosA)/1`
= sin A cos A
∴ L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/min.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
