SinA. CosA =...................
Anonymous:
answer should be 1
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sinA×cosA= tanASinA∗cosA=12∗sin(2A)=12∗2tanA1+tan2ASinA∗cosA=12∗sin(2A)=12∗2tanA1+tan2A
SinA∗cosA=sinA∗sin(π2−A)=sinA∗sinBSinA∗cosA=sinA∗sin(π2−A)=sinA∗sinB
SinA∗sinB=12[cos(A−B)−cos(A+B)]SinA∗sinB=12[cos(A−B)−cos(A+B)]
SinA∗cosA=sinA∗sin(π2−A)=sinA∗sinBSinA∗cosA=sinA∗sin(π2−A)=sinA∗sinB
SinA∗sinB=12[cos(A−B)−cos(A+B)]SinA∗sinB=12[cos(A−B)−cos(A+B)]
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