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Simplify: tan-1 xy-tan-1 x-yx+y

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प्रश्न

Simplify: `tan^-1  x/y - tan^-1  (x - y)/(x + y)`

योग
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उत्तर

`tan^-1  x/y - tan^-1  (x - y)/(x + y) = tan^-1 [(x/y - (x - y)/(x + y))/(1 + (x/y)((x - y)/(x + y)))]`

= `tan^-1 [((x(x + y) - y(x - y))/(y(x + y)))/(1 + (x(x - y))/(y(x + y)))]`

= `tan^-1 [(((x^2 + xy - xy + y^2))/(y(x + y)))/((y(x + y) + x(x - y))/(y(x + y)))]`

= `tan^-1 [(x^2 + y^2)/(xy + y^2 + x^2 - xy)]`

= `tan^-1 [(x^2 + y^2)/(x^2 + y^2)]`

= `tan^-1(1) = pi/4`

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अध्याय 4: Inverse Trigonometric Functions - Exercise 4.5 [पृष्ठ १६६]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.5 | Q 8 | पृष्ठ १६६

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