मराठी

Abbatan-1(acosx-bsinxbcosx-asinx),-π2<x<π2 and ababtanx>-1

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प्रश्न

`tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx)), - pi/2 < x < pi/2` and `"a"/"b" tan x > -1`

बेरीज
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उत्तर

Let y = `tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx))`

⇒ y = `tan^-1 [(("a"cosx)/("b"cosx) - ("b"sinx)/("b"cosx))/(("b"cosx)/("b"cosx) + ("a"sinx)/("b"cosx))]`

⇒ y = `tan^-1 [("a"/"b" - tanx)/(1 + "a"/"b" tanx)]`

⇒ y = `tan^-1  "a"/"b" - tan^-1 (tanx)`   ....`[because tan^-1  ((x - y)/(1 + xy)) = tan^-1x - tan^-1 y]`

⇒ y = `tan^-1  "a"/"b" - x`

Differentiating both sides with respect to x

`"dy"/"dx" = "d"/"dx"(tan^-1  "a"/"b") - "d"/"dx"(x)` = 0 – 1 = – 1

Hence, `"dy"/"dx"` = – 1.

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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 40 | पृष्ठ ११०

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