Question

. solve this fast
please​

Answer 1

Answer:

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Answer 2

Solution

We have,

sin3A−cos3AsinA−cosA+sinA1+tan2A−−−−−−−−−√−2tanAcotA.

=(sin2A+cos2A+sinAcosA)+sinA|secA|−2

=1+sinAcosA−sinAcosA−2=−1.