Q. Evaluate :  3   sin   3 A   +   2   cos   5 A + 10 ° 3   tan   3 A   -   cos e c   5 A - 20 ° , when A = 10 ° .

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Here is the solution of your asked query:
3 sin 3A+2 cos 5A+103 tan 3A-cosec5A-20=3 sin 30+2 cos50+103 tan 30-cosec50-20                 {putting A=10°}=312+2 cos 603 13-cosec 30=32+221-2=5-2=-52    ANS...
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