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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`
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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