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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Show that tan 75° + cot 75° = 4

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प्रश्न

Show that tan 75° + cot 75° = 4

बेरीज
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उत्तर

tan 75° = tan(45° + 30°)

= `(tan45^circ +  tan30^circ)/(1 - tan45^circ tan30^circ)`

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

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

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

cot 75° = `1/tan75^circ`

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

So, L.H.S = tan 75° + cot 75°

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

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

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

= `8/(3 - 1)`

= `8/2`

= 4

= R.H.S

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Trigonometric Functions and Their Properties
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.4 [पृष्ठ १०९]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.4 | Q 13 | पृष्ठ १०९

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