Question : The expression $\frac{\tan^{6} θ - \sec^{6} θ + 3 \sec^{2} θ \tan^{2} θ}{\tan^{2} θ + \cot^{2} θ + 2}, 0^{\circ} < θ < 90^{\circ}$ , is equal to:
Option 1: $\sec^{2} θ {cosec}^{2} θ$
Option 2: $- \sec^{2} θ {cosec}^{2} θ$
Option 3: $\cos^{2} θ \sin^{2} θ$
Option 4: $- \cos^{2} θ \sin^{2} θ$

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Question : $(\frac{\tan^{3} θ}{\sec^{2} θ} + \frac{\cot^{3} θ}{{cosec}^{2} θ} + 2 \sin θ \cos θ) \div (1 + {cosec}^{2} θ + \tan^{2} θ), 0^{\circ} < θ < 90^{\circ}$ , is equal to:

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Answer 1

Correct Answer: $- \cos^{2} θ \sin^{2} θ$

Solution : $\frac{\tan^{6} θ - \sec^{6} θ + 3 \sec^{2} θ \tan^{2} θ}{\tan^{2} θ + \cot^{2} θ + 2}$
Since $0^{\circ} < θ < 90^{\circ}$ , putting $θ = 45^{\circ}$
$\frac{\tan^{6} 45^{\circ} - \sec^{6} 45^{\circ} + 3 \sec^{2} 45^{\circ} \tan^{2} 45^{\circ}}{\tan^{2} 45^{\circ} + \cot^{2} 45^{\circ} + 2}$
$= \frac{1 - (\sqrt{2})^{6} + 3 \times (\sqrt{2})^{2} \times 1}{1 + 1 + 2}$
$= \frac{1 - 8 + 6}{1 + 1 + 2} = - \frac{1}{4}$
From option (iv),
$- \cos^{2} θ \sin^{2} θ$
Putting $θ = 45^{\circ}$ ,
$= - (\frac{1}{\sqrt{2}})^{2} (\frac{1}{\sqrt{2}})^{2} = - \frac{1}{4}$
Hence, the correct answer is $- \cos^{2} θ \sin^{2} θ$ .

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Question : The expression tan6⁡θ−sec6⁡θ+3sec2⁡θtan2⁡θtan2⁡θ+cot2⁡θ+2,0∘<θ<90∘, is equal to:Option 1: sec2⁡θcosec2θOption 2: −sec2⁡θcosec2θOption 3: cos2⁡θsin2⁡θOption 4: −cos2⁡θsin2⁡θ

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