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

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प्रश्न

Prove that:

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

योग

उत्तर

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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अध्याय 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (A) [पृष्ठ २९१]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (A) | Q 3.5 | पृष्ठ २९१

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