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प्रश्न
If θ < 90°, find the value of: sin2θ + cos2θ
उत्तर
Since θ <90°,
Consider θ = 45°
∴ sin2 + cos2
= sin245° + cos245°
= `(1/sqrt(2))^2 + (1/sqrt(2))^2`
= `(1)/(2) + (1)/(2)`
= 1.
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