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Question

In a ABC, if (sinA+cosA)(sinB+cosB)=2, then the value of 1+sec2(C2) is

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Solution

(sinA+cosA)(sinB+cosB)=2sinAsinB+sinAcosB +cosAsinB+cosAcosB=2cosAcosB+sinAsinB +cosAsinB+sinAcosB=2cos(AB)+sin(A+B)=2
We know that the maximum value of sine and cosine functions is 1, so
cos(AB)=1 and sin(A+B)=1AB=0 and A+B=π2A=B=π4C=π21+sec2(C2)=1+sec2(π4) =3

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