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Prove that: ((tan⁡60∘ +1)/(tan⁡60∘ –1))^2 =(1+cos⁡30∘)/(1–cos⁡30∘) - Mathematics

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Question

Prove that:

`((tan 60^circ  + 1)/(tan 60^circ  – 1))^2 = (1+ cos 30^circ) /(1– cos 30^circ) `

Sum
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Solution

LHS = `((tan 60^circ  + 1)/(tan 60^circ  – 1))^2`

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

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

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

= `(3 + 1 + 2 sqrt 3)/(3 + 1 - 2 sqrt 3)`

= `(4 + 2 sqrt 3)/(4 - 2 sqrt 3)`

= `(2(2 + sqrt 3))/(2(2 - sqrt 3)`

= `(2 + sqrt 3)/(2 - sqrt 3)`

R.H.S

= `(1 + cos 30^circ) /(1 - cos 30^circ)` 

= `(1 + sqrt 3/2)/(1 - sqrt 3/2)`

= `((2 + sqrt 3)/2)/((2 - sqrt 3)/2)`

= `(2 + sqrt 3)/(2 - sqrt 3)`

L.H.S = R.H.S

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Chapter 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (A) [Page 291]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (A) | Q 3.5 | Page 291
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