# Prove that: ( tan 60 ° + 1 tan 60 ° – 1 ) 2 = 1 + cos 30 ° 1 – cos 30 ° - Mathematics

Sum

Prove that:

((tan60°  + 1)/(tan 60°  – 1))^2 = (1+ cos 30°) /(1– cos 30°)

#### Solution

LHS = ((tan60°+ 1)/(tan 60° – 1))^2

= ((sqrt3 +1)/(sqrt3 - 1))^2

= (sqrt3 +1)^2/(sqrt3 -1)^2

= ((sqrt3)^2+(1)^2+2xxsqrt3xx1)/((sqrt3)^2+(1)^2-2xxsqrt3xx1)

= (3+1+2sqrt3)/(3+1-2sqrt3)

= (4 + 2sqrt3)/(4 -2sqrt3 )

= (2(2+sqrt3))/(2(2- sqrt3)

= (2+sqrt3)/(2-sqrt3)

R.H.S

= (1+ cos 30°) /(1- cos 30°)

= (1+sqrt3/2)/(1-sqrt3/2)

= ((2 + sqrt3)/2)/((2 - sqrt3)/2)

= (2+sqrt3)/(2-sqrt3)

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]

#### APPEARS IN

Selina Concise Mathematics 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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