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प्रश्न
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
पर्याय
\[\frac{16}{625}\]
\[\frac{1}{36}\]
\[\frac{3}{160}\]
\[\frac{160}{3}\]
उत्तर
Given: cos θ = 3/5 and we need to find the value of the following expression` "sinθ tanθ-1"/"2tan^2 θ"`
We know that `cos θ = "Base"/"Hypotenuse"`
⇒`"Base"=3 `
⇒ `"Hypotenuse"=5`
⇒`" Perpendicular"= sqrt(("Hypotenuse")^2-("Base")^2)`
⇒ `"Perpendicular"= sqrt(25-9)`
⇒`"Perpendicular"=4`
`"Since" sin θ= "Perpendicular"/"Hypotenuse"`
and tan θ= `"Perpendicular"/"Base" `
So we find,
`(sin θ tan θ-1)/(2 tan^2 θ)`
`(4/5xx4/3-1)/(2xx(4/3)^2)`
`(16/15-1)/(32/9)`
`(1/15)/(32/9)`
`3/160`
Hence the correct option is (c78)
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