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Q.

If k=(1+sinA)(1+sinB)(1+sinC)=(1sinA)(1sinB)(1sinC) then k=

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a

0

b

1

c

±cosAcosBcosC

d

±sinAsinBsinC

answer is B.

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Detailed Solution

k=(1+sinA)(1+sinB)(1+sinC)(1)

k=(1sinA)(1sinB)(1sinC)(2)

Eqn.(1) and Eqn.(2)

k2=(1sin2A)(1sin2B)(1sin2C)

=cos2A.cos2B.cos2C

k=±cosA.cosB.cosC

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