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प्रश्न
If cos θ = `12/13`, 0 < θ < `pi/2`, find the value of `(sin^2theta - cos^2theta)/(2sinthetacostheta), 1/(tan^2theta)`
उत्तर
cos θ = `12/13`
We know that,
sin2 θ = 1 – cos2θ
= `1 - (12/13)^2`
= `1 - 144/169`
= `25/169`
∴ sin θ = `± 5/13`
Since 0 < θ < `pi/2`,
θ lies in the 1st quadrant.
∴ sin θ > 0
∴ sin θ = `5/13`
∴ `(sin^2theta - cos^2theta)/(2sinthetacostheta) = ((5/13)^2 - (12/13)^2)/(2(5/13)(12/13)`
= `(25/169 - 144/169)/(120/169)`
= `(-119/169)/(120/169)`
= `-119/120`
tan θ = `sin theta/cos theta`
= `(5/13)/(12/13)`
= `5/12`
∴ `1/tan^2 theta = 1/(5/12)^2`
= `1/(25/144)`
= `144/25`
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