Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Advertisements
उत्तर
We have to prove (sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
We know that `sec^2 theta A - tan^2 theta A = 1`
So, we have
(sec A + tan A - 1)(sec A - tan A + 1) = {sec A + (tan A - 1)}{sec A - (tan A - 1)}
`= sec^2 A - (tan A - 1)^2`
`= sec^2 A - (tan^2 A - 2 tan A + 1)`
`= (sec^2 A - tan^2 A) + 2 tan A - 1`
So we have
(sec A + tan A - 1)(sec A - tan A + 1) = 1 + tan A - 1
= 2 tan A
Hence proved.
APPEARS IN
संबंधित प्रश्न
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
