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`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
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LHS = `cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta `
=`( cot ^2 theta + (cosec theta + 1 ) ^2 ) / ((cosec theta +1) cot theta)`
=` ( cot ^2 + cosec ^2 theta + 2 cosec theta +1 )/( (cosec theta +1) cot theta)`
=`( cot ^2 theta + cosec ^2 theta +2cosec theta + cosec ^2 theta - cot^2 theta)/((cosec theta +1 ) cot theta)`
=` (2 cosec^2 theta + 2 cosec theta)/(( cosec theta +1 ) cot theta)`
=`(2 cosec theta ( cosec theta +1))/(( cosec theta +1 ) cot theta)`
=` (2 cosec theta)/(cot theta)`
=`2 xx 1/sin theta xx sin theta/ cos theta`
= 2 sec ЁЭЬГ
= RHS
Hence, LHS = RHS
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If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Prove that cot2θ × sec2θ = cot2θ + 1.
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square ...`[cos theta = 1/sectheta]`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
