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प्रश्न
If sin θ − cos θ = 0 then the value of sin4θ + cos4θ
पर्याय
1
\[- 1\]
\[\frac{1}{2}\]
\[\frac{1}{4}\]
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उत्तर
`bb(1/2)`
Explanation:
It is given that,
\[\sin\theta - \cos\theta = 0\]
\[ \Rightarrow \sin\theta = \cos\theta\]
\[ \Rightarrow \frac{\sin\theta}{\cos\theta} = 1\]
\[ \Rightarrow \tan\theta = 1\]
\[ \Rightarrow \tan\theta = \tan45°\]
\[ \Rightarrow \theta = 45°\]
\[\therefore \sin^4 \theta + \cos^4 \theta\]
\[ = \sin^4 45° + \cos^4 45°\]
\[ = \left( \frac{1}{\sqrt{2}} \right)^4 + \left( \frac{1}{\sqrt{2}} \right)^4 \]
\[ = \frac{1}{4} + \frac{1}{4}\]
\[ = \frac{1}{2}\]
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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`
