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प्रश्न
Express `tan^-1 [(cos x)/(1 - sin x)], - pi/2 < x < (3pi)/2` in the simplest form.
उत्तर
`tan^-1 [(cos x)/(1 - sin x)]`
`= tan^-1 [(cos^2 x/2 - sin^2 x/2)/(cos^2 x/2 + sin^2 x/2 - 2 sin x/2 cos x/2)]`
`= tan^-1 [(cos^2 x/2 - sin^2 x/2)/(cos x/2 - sin x/2)^2]`
`= tan^-1 [((cos x/2 - sin x/2)(cos x/2 + sin x/2))/(cos x/2 - sin x/2)^2]`
[∵ a2 – b2 = (a + b) (a – b)]
`= tan^-1 [(cos x/2 + sin x/2)/(cos x/2 - sin x/2)]`
`= tan^-1 [((cos x/2)/(cos x/2) + (sin x/2)/(cos x/2))/((cos x/2)/(cos x/2) - (sin x/2)/(cos x/2))]`
[∵ Divide each term by cos `x/2`]
`= tan^-1 [(1 + tan x/2)/(1 - tan x/2)]`
`= tan^-1 [(tan pi/4 + tan x/2)/(1 - tan pi/4 tan x/2)]`
`= tan^-1 [tan (pi/4 + x/2)] = pi/4 + x/2`
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