Solve the following equations:tanθ+tan(θ+π3)+tan(θ+2π3)=3

प्रश्न

Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`

बेरीज

उत्तर

Now `tan(theta + pi/3) = (tantheta + sqrt(3))/(1 - sqrt(3) tan theta)` and `tan(theta + (2pi)/3)`

= `(tantheta - sqrt(3))/(1 + sqrt(3) tan theta)`

So, `tan(theta + pi/3) + tan(theta + (2pi)/3)`

= `(tantheta + sqrt(3))/(1 - sqrt(3) tantheta) + (tan theta - sqrt(3))/(1 + sqrt(3) tantheta)`

= `((tan theta + sqrt(3))(1 + sqrt(3) tan theta) + (tan theta - sqrt(3))(1 - sqrt(3) tan theta))/(1 - 3tan^2theta)`

= `(tan theta + sqrt(3) + sqrt(3)tan^2theta + 3tantheta + tantheta - sqrt(3)tan^2theta - sqrt(3) + 3tantheta)/(1 - 3tan^2theta)`

= `(8tantheta)/(1 - 3tan^2theta)`

Given, `tantheta + tan(theta + pi/3) + tan(theta + (2pi)/3) = sqrt(3)`

⇒ `tan theta + (8tantheta)/(1 - 3tan^2theta) = sqrt(3)`

⇒ `(tantheta - 3tan^3theta + 8tantheta)/(1 - 3tan^2theta) = sqrt(3)`

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Trigonometric Equations

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 3. (ix)Q 3. (viii)Q 3. (x)

पाठ 3: Trigonometry - Exercise 3.8 [पृष्ठ १३३]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board

पाठ 3 Trigonometry
Exercise 3.8 | Q 3. (ix) | पृष्ठ १३३

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