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प्रश्न
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
उत्तर
If `sqrt(3)` sin θ – cos θ = θ
To prove tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
`sqrt(3)` sin θ – cos θ = θ
`sqrt(3)` sin θ = cos θ
`sin theta/cos theta = 1/sqrt(3)`
tan θ = tan 30°
θ = 30°
L.H.S = tan 3θ°
= tan3 (30°)
= tan 90°
= undefined (α)
R.H.S = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
= `(3tan30^circ - tan^2 30^circ)/(1 - 3tan^2 30^circ)`
= `3(1/sqrt(3)) - (1/sqrt(3))^3 ÷ 1 - 3 xx (1/sqrt(3))^2`
= `sqrt(3) - 1/(3sqrt(3)) ÷ 1 - 3 xx 1/3`
= `(9 - 1)/(3sqrt(3)) ÷ 1 - 1`
= `8/(3sqrt(3)) ÷ 0`
= undefined (α)
∴ tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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