Advertisements
Advertisements
Question
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
Advertisements
Solution
L.H.S. = `sqrt((1 + cos A)/(1 - cos A))`
= `sqrt((1 + cos A)/(1 - cos A) xx (1 + cos A)/(1 + cos A))` ...[On rationalising the denominator]
= `sqrt((1 + cos A)^2/(1 - cos^2 A))`
= `sqrt((1 + cos A)^2/(sin^2 A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= `(1 + cos A)/(sin A)`
= `1/(sin A) + (cos A)/(sin A)`
= cosec A + cot A
= R.H.S.
∴ `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Simplify : 2 sin30 + 3 tan45.
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
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)
cot θ . tan θ = ?
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
