Question

if sec theta = m+n/2√mn , then sin theta is
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Answer 1

Answer:

sec@=m+n/2√mn

cos@=2√mn/m+n

sin^2@=1-cos^2@

1-4mn/(m+n)^2

=(m+n)^2-4mn/(m+n)^2

=m^2+2mn+n^2-4mn/(m+n)^2

=m^2-2mn+n^2/(m+n)^2

=(m-n)^2/(m+n)^2

sin@=(m-n)/(m+n)