Let $$f($$θ$$)=tan3$$⁡θ$$1+tan2$$⁡θ−$$cot3$$⁡θ$$1+cot2$$⁡θ,$$0$$<θ<π$$4$$ Then f $$($$θ$$)$$ is equal to

Correct option is 3−2cot⁡(2θ)

Questions

Let $f (θ) = \frac{\tan^{3} θ}{1 + \tan^{2} θ} - \frac{\cot^{3} θ}{1 + \cot^{2} θ}, 0 < θ < \frac{π}{4}$ Then $f (θ)$ is equal to

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tan⁡θ+cot⁡θ

2sin⁡(2θ)

−2cot⁡(2θ)

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detailed solution

Correct option is C

f(θ)=sin3⁡θcos⁡θ−cos3⁡θsin⁡θ=sin4⁡θ−cos4⁡θcos⁡θsin⁡θ−−cos2⁡θ−sin2⁡θ(1)cos⁡θsin⁡θ=−2cot⁡(2θ)

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Let f(θ)=tan3⁡θ1+tan2⁡θ−cot3⁡θ1+cot2⁡θ,0<θ<π4 Then f (θ) is equal to

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Let $f (θ) = \frac{\tan^{3} θ}{1 + \tan^{2} θ} - \frac{\cot^{3} θ}{1 + \cot^{2} θ}, 0 < θ < \frac{π}{4}$ Then $f (θ)$ is equal to