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If tan x = nnnn+1 and tan y = n12n+1, find tan(x + y) - Mathematics

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प्रश्न

If tan x = `"n"/("n" + 1)` and tan y = `1/(2"n" + 1)`, find tan(x + y)

योग
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उत्तर

tan x = `"n"/("n" + 1)`, tan y = `1/(2"n" + 1)`

tan(x + y) = `(tanx + tany)/(1 - tanx tany)`

= `("n"/("n" + 1) + 1/(2"n" + 1))/(1 - "n"/("n" + 1)  * 1/(2"n" + 1))`

= `(("n"(2"n" + 1) + "n" + 1)/(("n" + 1)(2"n" + 1)))/((("n" + 1)(2"n" + 1) - "n")/(("n" + 1)(2"n" + 1))`

= `("n"(2"n" + 1) + "n" + 1)/(("n" + 1)(2"n" + 1) - "n")`

= `(2"n"^2 + "n" + "n" + 1)/(2"n"^2 + "n" + 2"n" + 1 - "n")`

= `(2"n"^2 + 2"n" + 1)/(2"n"^2 + 2"n" + 1)`

tan(x + y) = 1

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.4 [पृष्ठ ११०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.4 | Q 22 | पृष्ठ ११०

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