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Let $45^{\circ} \leq \theta < 90^{\circ}$. If $\tan \theta+\cot \theta=(\tan \theta)^{\mathrm{i}}+(\cot \theta)^{\mathrm{j}}$ for some
$\mathrm{i} \geq 2$, then what is the value of $\sin \theta+\cos \theta$ ?
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$\mathrm{i} \geq 2$, then what is the value of $\sin \theta+\cos \theta$ ?
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Verified Answer
The correct answer is:
$\sqrt{2}$
As given, $\tan \theta+\cot \theta=(\tan \theta)^{\mathrm{i}}+(\cot \theta)^{\mathrm{i}}$
Also, $45^{\circ} \leq \theta < 90^{\circ}$ and $\mathrm{i} \geq 2$
which is possible only when $\theta=45^{\circ}$ Since, $\tan 45^{\circ}+\cot 45^{\circ}=1+1=2$
and $\left(\tan 45^{\circ}\right)^{\mathrm{i}}+\left(\cot 45^{\circ}\right)^{\mathrm{i}}=1+1=2$
Thus, $\sin \theta+\cos \theta=\sin 45^{\circ}+\cos 45^{\circ}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
Also, $45^{\circ} \leq \theta < 90^{\circ}$ and $\mathrm{i} \geq 2$
which is possible only when $\theta=45^{\circ}$ Since, $\tan 45^{\circ}+\cot 45^{\circ}=1+1=2$
and $\left(\tan 45^{\circ}\right)^{\mathrm{i}}+\left(\cot 45^{\circ}\right)^{\mathrm{i}}=1+1=2$
Thus, $\sin \theta+\cos \theta=\sin 45^{\circ}+\cos 45^{\circ}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
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