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Question

Simplify:
tan1[2cosx3sinx3cosx+2sinx],where 23tanx>1

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Solution

We have, tan1[2cosx3sinx3cosx+2sinx]

Let a=tan1[2cosx3sinx3cosx+2sinx]

=tan1⎢ ⎢ ⎢ ⎢2cosx(132sinxcosx)3cosx(1+23sinxcosx)⎥ ⎥ ⎥ ⎥

=tan1⎢ ⎢ ⎢23(132tanx)(1+23tanx)⎥ ⎥ ⎥

=tan1⎢ ⎢ ⎢23×32⎢ ⎢ ⎢23tanx1+23tanx⎥ ⎥ ⎥⎥ ⎥ ⎥

=tan1⎢ ⎢ ⎢23tanx1+23tanx⎥ ⎥ ⎥

Let 23=tanα,α=tan1=23

=tan1[tanαtanx1+tanαtanx]

=tan1tan(αx)

=αx

a=tan123x

Hence, this is the answer.

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