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Question
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
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Solution
LHS = `tan^3 θ/(1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ)`
= `tan^3 θ/sec^2 θ + cot^3 θ/(cosec^2 θ)`
= 1 + tan2θ = sec2θ; 1 + cot2θ = cosec2θ
= `sin^3 θ/cos^3 θ xx cos^2 θ + cos^3 θ/sin^3 θ xx sin^2 θ`
= `sin^3 θ/cos θ + cos^3 θ/sin θ`
= `(sin^4 θ + cos^4 θ)/(cos θ.sin θ)`
= `((sin^2θ)^2 + (cos^2θ)^2)/(sin θ.cos θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ. cos^2 θ)/(sin θ.cos θ)` ...[a2 + b2 = (a + b)2 − 2ab]
= `((1)^2 - 2sin^2θ. cos^2θ)/(sin θ.cos θ)`
= `(1 - 2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= `1/(sinθ.cosθ) - (2sin^2θ. cos^2θ)/(sinθ.cosθ)`
= secθ. cosecθ − 2 sinθ cosθ
= RHS
Hence proved.
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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]`
