Advertisements
Advertisements
Question
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Solution
L.H.S
`(1+secA)/secA = (1+1/(cosA))/(1/cosA)`
= `((cosA+1)/cosA)/(1/cosA)`
= `(cosA+1)`
= `((1-cosA)(1+cosA))/(1-cosA)`
= `(1-cos^2A)/(1-cosA)`
= `(sin^2A)/(1-cosA)` ...[∵ 1cos2 A = sin2A]
R.H.S
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`
If `sin theta = x , " write the value of cot "theta .`
Define an identity.
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2cosecθ`
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Choose the correct alternative:
cot θ . tan θ = ?
Choose the correct alternative:
sec2θ – tan2θ =?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
If 5 tan β = 4, then `(5 sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
