Question

If $\dfrac{\pi }{2} < x < \pi $ , then $\sqrt{\dfrac{1-\sin x}{1+\sin x}}+\sqrt{\dfrac{1+\sin x}{1-\sin x}}$ is equal to
(A) $2\sec x$
(B) $-2\sec x$
(C) $\sec x$
(D) $-\sec x$

Answer 1

Hint: At first the simplify the two terms by multiplying $\sqrt{\left( 1-\sin x \right)}$ in the numerator and denominator of the first term and $\sqrt{\left( 1+\sin x \right)}$ in the numerator and denominator of the second term. The given condition is that $\dfrac{\pi }{2} < x < \pi $ , so, $\sqrt{{{\cos }^{2}}x}$ , must be $-\cos x$ . Having done all of these, we end up with $-2\sec x$ as our solution.

Complete step-by-step answer:
The given expression is
$\sqrt{\dfrac{1-\sin x}{1+\sin x}}+\sqrt{\dfrac{1+\sin x}{1-\sin x}}$
We rewrite the term $\sqrt{\dfrac{1-\sin x}{1+\sin x}}$ as $\sqrt{\dfrac{\left( 1-\sin x \right)\times \left( 1-\sin x \right)}{\left( 1+\sin x \right)\times \left( 1-\sin x \right)}}$ by multiplying $\sqrt{\left( 1-\sin x \right)}$ in the numerator and denominator both. The terms thus becomes,
$\Rightarrow \sqrt{\dfrac{{{\left( 1-\sin x \right)}^{2}}}{1-{{\sin }^{2}}x}}$
Now, $1-{{\sin }^{2}}x$ can be written as ${{\cos }^{2}}x$ as ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . The term thus becomes
$\Rightarrow \sqrt{\dfrac{{{\left( 1-\sin x \right)}^{2}}}{{{\cos }^{2}}x}}$
Applying the square root to the squares, the term thus becomes,
$\Rightarrow \dfrac{\left( 1-\sin x \right)}{\cos x}$
But, as $\dfrac{\pi }{2} < x < \pi $ , this means $-1 < \cos x < 0$ or that $\cos x$ is negative in the region $\left( \dfrac{\pi }{2},\pi \right)$ . In general, $\sqrt{x}$ must always be positive. Therefore, the expression instead of being $\dfrac{\left( 1-\sin x \right)}{\cos x}$ , must be
$\Rightarrow \dfrac{\left( 1-\sin x \right)}{-\cos x}....term1$
We do the same thing with the second term, but here we multiply $\sqrt{\left( 1+\sin x \right)}$ in the numerator and denominator both. The terms thus becomes,
$\Rightarrow \sqrt{\dfrac{\left( 1+\sin x \right)\times \left( 1+\sin x \right)}{\left( 1-\sin x \right)\times \left( 1+\sin x \right)}}$
The term thus becomes,
$\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin x \right)}^{2}}}{1-{{\sin }^{2}}x}}$
Now, $1-{{\sin }^{2}}x$ can be written as ${{\cos }^{2}}x$ as ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . The term thus becomes
$\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin x \right)}^{2}}}{{{\cos }^{2}}x}}$
Applying the square root to the squares, the term thus becomes,
$\Rightarrow \dfrac{\left( 1-\sin x \right)}{\cos x}$
But, as $\dfrac{\pi }{2} < x < \pi $ , this means $-1 < \cos x < 0$ or that $\cos x$ is negative in the region $\left( \dfrac{\pi }{2},\pi \right)$ . In general, $\sqrt{x}$ must always be positive. Therefore, the expression instead of being $\dfrac{\left( 1-\sin x \right)}{\cos x}$ , must be
$\Rightarrow \dfrac{\left( 1+\sin x \right)}{-\cos x}....term2$
Adding $term1$ and $term2$ , we get,
$\dfrac{\left( 1-\sin x \right)}{-\cos x}+\dfrac{\left( 1+\sin x \right)}{-\cos x}$
The denominators being the same, we can directly add the numerators. The expression thus becomes
$\Rightarrow \dfrac{\left( 1-\sin x \right)+\left( 1+\sin x \right)}{-\cos x}$
$\sin x$ terms get cancelled out in the numerator and we are left with,
$\Rightarrow \dfrac{2}{-\cos x}$
We can write $\dfrac{1}{\cos x}$ as $\sec x$ . the expression thus becomes,
$\Rightarrow -2\sec x$
Therefore, we can conclude that the given expression is equal to $-2\sec x$ or option B.
So, the correct answer is “Option B”.

Note: The problem may seem easy at first and it really is if we don’t overlook the given condition that $\dfrac{\pi }{2} < x < \pi $ . Most of us overlook this and assume $x$ to be in the interval $\left( 0,\dfrac{\pi }{2} \right)$ where $\cos x$ is positive so, taking roots would not create any problem. The simplification of the individual terms must be done carefully.