In a right angled triangle ABC, write the value of $s i n^{2} A + s i n^{2} B + s i n^{2} C$ .

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Solution

Suppose in $Δ A B C ∠ B = 90^{\circ}$

$\Rightarrow A + C = \frac{π}{2}$

$\Rightarrow A = \frac{π}{2} - C$

$\Rightarrow s i n A = s i n (\frac{π}{2} - C) = c o s C$ Now, $s i n^{2} A + s i n^{2} B + s i n^{2} C$

$= s i n^{2} A + 1 + c o s^{2} A [∵ s i n \frac{π}{2} = 1]$

= 1 + 1 = 2

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