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Question
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
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Solution
L.H.S. = tan2Φ + cot2Φ + 2
= tan2Φ + 1 + cot2Φ + 1
= sec2Φ + cosec2Φ
= `1/cos^2 Φ + 1/sin^2Φ`
= `(sin^2 Φ + cos^2 Φ)/(sin^2 Φ.cos^2Φ )`
= `1/(sin^2 Φ. cos^2 Φ )`
= cosec2Φ. sec2Φ
= R.H.S.
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
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]`
