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If-tan-cos-9-sin-9-cos-9-sin-9-then-




Question Number 49459 by Pk1167156@gmail.com last updated on 07/Dec/18
If tan θ=((cos 9°+sin 9°)/(cos 9°−sin 9°)) then θ = ___
$$\mathrm{If}\:\mathrm{tan}\:\theta=\frac{\mathrm{cos}\:\mathrm{9}°+\mathrm{sin}\:\mathrm{9}°}{\mathrm{cos}\:\mathrm{9}°−\mathrm{sin}\:\mathrm{9}°}\:\mathrm{then}\:\theta\:=\:\_\_\_ \\ $$
Answered by math1967 last updated on 07/Dec/18
tanθ=((cos9+cos(90−9))/(cos9−cos(90−9)))  tanθ=((cos9+cos81)/(cos9−cos81))  tanθ=((2cos45.cos36)/(2sin45.sin36))  tanθ=cot36    [∵cos45=sin45]  tanθ=cot36  tanθ=tan(90−36)  tanθ=tan54 ∴θ=54 ans
$${tan}\theta=\frac{{cos}\mathrm{9}+{cos}\left(\mathrm{90}−\mathrm{9}\right)}{{cos}\mathrm{9}−{cos}\left(\mathrm{90}−\mathrm{9}\right)} \\ $$$${tan}\theta=\frac{{cos}\mathrm{9}+{cos}\mathrm{81}}{{cos}\mathrm{9}−{cos}\mathrm{81}} \\ $$$${tan}\theta=\frac{\mathrm{2}{cos}\mathrm{45}.{cos}\mathrm{36}}{\mathrm{2}{sin}\mathrm{45}.{sin}\mathrm{36}} \\ $$$${tan}\theta={cot}\mathrm{36}\:\:\:\:\left[\because{cos}\mathrm{45}={sin}\mathrm{45}\right] \\ $$$${tan}\theta={cot}\mathrm{36} \\ $$$${tan}\theta={tan}\left(\mathrm{90}−\mathrm{36}\right) \\ $$$${tan}\theta={tan}\mathrm{54}\:\therefore\theta=\mathrm{54}\:{ans} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
tanθ=((1+tan9^o )/(1−tan9^o ))=tan(45^o +9^o )=tan54^o
$${tan}\theta=\frac{\mathrm{1}+{tan}\mathrm{9}^{{o}} }{\mathrm{1}−{tan}\mathrm{9}^{{o}} }={tan}\left(\mathrm{45}^{{o}} +\mathrm{9}^{{o}} \right)={tan}\mathrm{54}^{{o}} \\ $$

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