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प्रश्न
The value of sin2 29° + sin2 61° is
विकल्प
1
0
2 sin2 29°
2 cos2 61°
उत्तर
The given expression is `sin^29°+sin^2 61°`
`sin^2 29°+sin^2 61°`.
=` sin^2 29°+(sin 61°)^2`
`= sin^2 29°+{sin(90°-29°)}^2`
`=sin^2 29°+(cos 29°)^2`
`= sin^2 29°+cos^2°29°`
`= 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`
