Let S=50tan(3tan−121+2cos−151)+42tan(21tan−122)
=50tan(tan−121+2(tan−121+tan−12))+42tan(21tan−122)
Let 21tan−122=α⇒tan2α=22
Now 1−tan2α2tanα=22
Solving we get, tanα=21
SoS=50tan(tan−121+2⋅2π))+42×21
=50(tantan−121)+4
=25+4=29
50tan(3tan−1(21)+2cos−1(51))+42tan(21tan−1(22)) is equal to ______.
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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