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प्रश्न
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
उत्तर
Given that,
`sqrt(3) tan θ` = 1
⇒ tan θ = `1/sqrt(3)` = tan 30°
⇒ θ = 30°
Now, sin2θ – cos2θ = sin230° – cos230°
= `(1/2)^2 - (sqrt(3)/2)^2`
= `1/4 - 3/4`
= `(1 - 3)/4`
= `-2/4`
= `-1/2`
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Activity:
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= `square (1 - (sin^2theta)/(tan^2theta))`
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