In a right angled triangle ABC, ∠B=90o. If sinA=√32 then verifying the following results: (i) sin(A+C)=sinAcosC+cosA.sinC (ii) cos(A+C)=cosAcosC−sinAsinC
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Solution
From the identity, sin2A+cos2A=1, and given sinA=√32, we obtain cosA=12, given ∠B=90o,weobtain∠A=sin−1√32=60o
Hence ∠C=30o.
By substituting the obtained values in (i), sin(90o)=√32.√32+12.12
1=1
For verifying (ii), substitute values in (ii), cos(90o)=12.√32−√32.12