Question

Prove that $\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} \right|=\dfrac{-2}{\cos \theta },\text{ where }\dfrac{\pi }{2}<\theta <\pi $

Answer 1

Hint: Simply the expression $\dfrac{1-\sin \theta }{1+\sin \theta }$ by multiplying the numerator and denominator by $1-\sin \theta $ and using the identities $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $. Similarly, simplify the expression $\dfrac{1+\sin \theta }{1-\sin \theta }$ by multiplying the numerator and the denominator by $1+\sin \theta $ and using the identities $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $. Hence find the simplified expressions for $\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}$ and $\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}$. Hence evaluate the expression \[\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} \right|\] and hence prove L.H.S = R.H.S.

Complete step-by-step answer:
Simplifying the expression $\dfrac{1-\sin \theta }{1+\sin \theta }$:
Multiplying the numerator and denominator by $1-\sin \theta $, we get
$\dfrac{1-\sin \theta }{1+\sin \theta }=\dfrac{1-\sin \theta }{1+\sin \theta }\times \dfrac{1-\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above algebraic identity, we get
$\dfrac{1-\sin \theta }{1+\sin \theta }=\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{1-{{\sin }^{2}}\theta }$
We know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $
Hence, we have \[\dfrac{1-\sin \theta }{1+\sin \theta }=\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }\]
Simplifying the expression $\dfrac{1+\sin \theta }{1-\sin \theta }$:
Multiplying the numerator and denominator by $1+\sin \theta $, we get
$\dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{1+\sin \theta }{1-\sin \theta }\times \dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above algebraic identity, we get
$\dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{1-{{\sin }^{2}}\theta }$
We know that $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $
Hence, we have \[\dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }\]
Hence, we have
$\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}$
We know that $\sqrt{{{x}^{2}}}=\left| x \right|$
Hence, we have
$\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=\left| \dfrac{1-\sin \theta }{\cos \theta } \right|$
Similarly, we have $\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}=\sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}=\left| \dfrac{1+\sin \theta }{\cos \theta } \right|$
We know that $-1\le \sin \theta \le 1$
Hence, we have $1+\sin \theta \ge 0,1-\sin \theta \ge 0$
Also, since $\dfrac{\pi }{2}<\theta <\pi ,$ we have $\cos \theta <0$
Hence, we have $\left| \cos \theta \right|=-\cos \theta ,\left| 1+\sin \theta \right|=1+\sin \theta ,\left| 1-\sin \theta \right|=1-\sin \theta $
Hence, we have
$\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} \right|=\left| \left| \dfrac{1-\sin \theta }{\cos \theta } \right|+\left| \dfrac{1+\sin \theta }{\cos \theta } \right| \right|$
Since $\left| \cos \theta \right|=-\cos \theta ,\left| 1+\sin \theta \right|=1+\sin \theta ,\left| 1-\sin \theta \right|=1-\sin \theta $ we have
$\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} \right|=\left| -\dfrac{1-\sin \theta }{\cos \theta }-\dfrac{1+\sin \theta }{\cos \theta } \right|=\left| \dfrac{2}{\cos \theta } \right|$
Since $\left| \cos \theta \right|=-\cos \theta ,$ we have
$\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} \right|=\dfrac{-2}{\cos \theta }$

Note: [1] Many students make a mistake by writing $\sqrt{{{\cos }^{2}}x}=\cos x$. This is incorrect as the square root (L.H.S) is a positive quantity, whereas cosx (R.H.S) can be positive as well as a negative quantity. In general, we have $\sqrt{{{x}^{2}}}=\left| x \right|$ and not $\sqrt{{{x}^{2}}}=x$