Math, asked by gyanchandinisha8, 18 days ago

tan^dita (1+cot^ dota)

Answers

Answered by daseshan660

0

Answer:

LHS

1−cotθtanθ+1−tanθcotθ

=1−tanθ1tanθ+1−tanθtanθ1

=tanθtanθ−1tanθ+tanθ(1−tanθ)1

=tanθ−1tan2θ+tanθ(1−tanθ)1

=tanθ−1tan2θ−tanθ(tanθ−1)1

=tanθ(tanθ−1)tan3θ−1

=tanθ(tanθ−1)(tanθ−1)(tan2θ+tanθ+1)

=tanθtan2θ+tanθ+1

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