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`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
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LHS = `(1- tan^2 theta)/(cot^2 theta-1)`
=`(1-(sin^2 theta)/(cos^2 theta))/((cos^2 theta )/(sin^2 theta)-1)`
=`((cos^2 theta - sin^2 theta)/(cos^2 theta))/((cos^2theta-sin^2 theta)/(sin^2 theta))`
=`(sin^2 theta)/(cos^2 theta)`
= tan2 ЁЭЬГ
= RHS
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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.]
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
sec4 A − sec2 A is equal to
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
If sin θ = `1/2`, then find the value of θ.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
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 that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
