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If $\tan \theta=\sqrt{m}$, where $\mathrm{m}$ is non-square natural
number, then $\sec 2 \theta$ is
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number, then $\sec 2 \theta$ is
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Verified Answer
The correct answer is:
a negative number
Let $\tan \theta=\sqrt{m}$, where $\mathrm{m}$ is a non-square natural num-
ber.
$\Rightarrow \sin \theta=\sqrt{m} \cos \theta$
Consider, $\sec 2 \theta=\frac{1}{\cos 2 \theta}=\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta-m \cos ^{2} \theta}=\frac{1}{\cos ^{2} \theta(1-m)}$
$=\frac{\sec ^{2} \theta}{1-m}=\frac{1+\tan ^{2} \theta}{1-m}=\frac{1+m}{1-m}$
$=\frac{(1+m)(1-m)}{(1-m)(1-m)}=\frac{\left(1-m^{2}\right)}{(1-m)^{2}}$
Numerator will always be negative and denominator will always be positive
Hence, $\sec 2 \theta=\frac{1-m^{2}}{(1-m)^{2}}$ is a negative number
ber.
$\Rightarrow \sin \theta=\sqrt{m} \cos \theta$
Consider, $\sec 2 \theta=\frac{1}{\cos 2 \theta}=\frac{1}{\cos ^{2} \theta-\sin ^{2} \theta}$
$=\frac{1}{\cos ^{2} \theta-m \cos ^{2} \theta}=\frac{1}{\cos ^{2} \theta(1-m)}$
$=\frac{\sec ^{2} \theta}{1-m}=\frac{1+\tan ^{2} \theta}{1-m}=\frac{1+m}{1-m}$
$=\frac{(1+m)(1-m)}{(1-m)(1-m)}=\frac{\left(1-m^{2}\right)}{(1-m)^{2}}$
Numerator will always be negative and denominator will always be positive
Hence, $\sec 2 \theta=\frac{1-m^{2}}{(1-m)^{2}}$ is a negative number
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