In a right-angled triangle ABC, $∠ B = 90^{o}$ then find the value of $sin A . cos C + cos A . sin C$ .

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Solution

Given in $△$ ABC, $∠ B = 90^{o}$ .
This given $A + C = 180^{o} - 90^{o} = 90^{o}$ ......(1).
Now,
$sin A . cos C + cos A . sin C$
$= sin (A + C)$
$= sin (90^{o})$
$= 1$ .

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