cosec(65°+A)-sec (25°-A)
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5
cosec(65°+A)-sec(25-A)
= cosec(65°+A)-sec[90°-(65°-A)]
=cosec(65°+A)- cosec(65°+A)=0
therefore ans =0
= cosec(65°+A)-sec[90°-(65°-A)]
=cosec(65°+A)- cosec(65°+A)=0
therefore ans =0
Answered by
4
cosec (65°+A)-sec (25°-A)
sec{90-(65+A)}-sec(25-A)
sec(25+A)- sec (25-A)
0
sec{90-(65+A)}-sec(25-A)
sec(25+A)- sec (25-A)
0
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