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प्रश्न
Evaluate sin25° cos65° + cos25° sin65°
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उत्तर
sin25° cos65° + cos25° sin65°
=(sin 25°) {cos(90°-25°)}+cos 25°{sin(90°-25)}
=(sin 25°)(sin 25°) + (cos 25°)(cos 25°)
= sin225° + cos225°
= 1 (As sin2A + cos2A = 1)
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
