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Question: Prove that \(\left| \sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1+\sin \theta }{1-\si...

Prove that ∣1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ∣=−2cos⁡θ, where π2<θ<π\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

Explanation

Solution

Hint: Simply the expression 1−sin⁡θ1+sin⁡θ\dfrac{1-\sin \theta }{1+\sin \theta } by multiplying the numerator and denominator by 1−sin⁡θ1-\sin \theta and using the identities (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta . Similarly, simplify the expression 1+sin⁡θ1−sin⁡θ\dfrac{1+\sin \theta }{1-\sin \theta } by multiplying the numerator and the denominator by 1+sin⁡θ1+\sin \theta and using the identities (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} and 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta . Hence find the simplified expressions for 1−sin⁡θ1+sin⁡θ\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }} and 1+sin⁡θ1−sin⁡θ\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}. Hence evaluate the expression ∣1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ∣\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 1−sin⁡θ1+sin⁡θ\dfrac{1-\sin \theta }{1+\sin \theta }:
Multiplying the numerator and denominator by 1−sin⁡θ1-\sin \theta , we get
1−sin⁡θ1+sin⁡θ=1−sin⁡θ1+sin⁡θ×1−sin⁡θ1−sin⁡θ=(1−sin⁡θ)2(1−sin⁡θ)(1+sin⁡θ)\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 (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
Using the above algebraic identity, we get
1−sin⁡θ1+sin⁡θ=(1−sin⁡θ)21−sin⁡2θ\dfrac{1-\sin \theta }{1+\sin \theta }=\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{1-{{\sin }^{2}}\theta }
We know that 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta
Hence, we have 1−sin⁡θ1+sin⁡θ=(1−sin⁡θ)2cos⁡2θ\dfrac{1-\sin \theta }{1+\sin \theta }=\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }
Simplifying the expression 1+sin⁡θ1−sin⁡θ\dfrac{1+\sin \theta }{1-\sin \theta }:
Multiplying the numerator and denominator by 1+sin⁡θ1+\sin \theta , we get
1+sin⁡θ1−sin⁡θ=1+sin⁡θ1−sin⁡θ×1+sin⁡θ1−sin⁡θ=(1+sin⁡θ)2(1−sin⁡θ)(1+sin⁡θ)\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 (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
Using the above algebraic identity, we get
1+sin⁡θ1−sin⁡θ=(1+sin⁡θ)21−sin⁡2θ\dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{1-{{\sin }^{2}}\theta }
We know that 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta
Hence, we have 1+sin⁡θ1−sin⁡θ=(1+sin⁡θ)2cos⁡2θ\dfrac{1+\sin \theta }{1-\sin \theta }=\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }
Hence, we have
1−sin⁡θ1+sin⁡θ=(1−sin⁡θ)2cos⁡2θ\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}
We know that x2=∣x∣\sqrt{{{x}^{2}}}=\left| x \right|
Hence, we have
1−sin⁡θ1+sin⁡θ=∣1−sin⁡θcos⁡θ∣\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=\left| \dfrac{1-\sin \theta }{\cos \theta } \right|
Similarly, we have 1+sin⁡θ1−sin⁡θ=(1+sin⁡θ)2cos⁡2θ=∣1+sin⁡θcos⁡θ∣\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≤sin⁡θ≤1-1\le \sin \theta \le 1
Hence, we have 1+sin⁡θ≥0,1−sin⁡θ≥01+\sin \theta \ge 0,1-\sin \theta \ge 0
Also, since π2<θ<π,\dfrac{\pi }{2}<\theta <\pi , we have cos⁡θ<0\cos \theta <0
Hence, we have ∣cos⁡θ∣=−cos⁡θ,∣1+sin⁡θ∣=1+sin⁡θ,∣1−sin⁡θ∣=1−sin⁡θ\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
∣1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ∣=∣∣1−sin⁡θcos⁡θ∣+∣1+sin⁡θcos⁡θ∣∣\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 ∣cos⁡θ∣=−cos⁡θ,∣1+sin⁡θ∣=1+sin⁡θ,∣1−sin⁡θ∣=1−sin⁡θ\left| \cos \theta \right|=-\cos \theta ,\left| 1+\sin \theta \right|=1+\sin \theta ,\left| 1-\sin \theta \right|=1-\sin \theta we have
∣1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ∣=∣−1−sin⁡θcos⁡θ−1+sin⁡θcos⁡θ∣=∣2cos⁡θ∣\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 ∣cos⁡θ∣=−cos⁡θ,\left| \cos \theta \right|=-\cos \theta , we have
∣1−sin⁡θ1+sin⁡θ+1+sin⁡θ1−sin⁡θ∣=−2cos⁡θ\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 cos⁡2x=cos⁡x\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 x2=∣x∣\sqrt{{{x}^{2}}}=\left| x \right| and not x2=x\sqrt{{{x}^{2}}}=x