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Question: Prove the following statement: \[\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A\]...

Prove the following statement:
1−sin⁡A1+sin⁡A=sec⁡A−tan⁡A\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A

Explanation

Solution

Hint: First of all consider the LHS of the given equation and multiply it by 1−sin⁡A1−sin⁡A\sqrt{\dfrac{1-\sin A}{1-\sin A}}. Now, use 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta and simplify the given expression. Now, separate the terms to prove the desired result.
Complete step-by-step answer:
Here, we have to prove that 1−sin⁡A1+sin⁡A=sec⁡A−tan⁡A\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A. Let us consider the LHS of the equation given in the question.
E=1−sin⁡A1+sin⁡AE=\sqrt{\dfrac{1-\sin A}{1+\sin A}}
By multiplying the above expression by 1−sin⁡A1−sin⁡A\sqrt{\dfrac{1-\sin A}{1-\sin A}}, we get,
E=1−sin⁡A1+sin⁡A.1−sin⁡A1−sin⁡AE=\sqrt{\dfrac{1-\sin A}{1+\sin A}}.\sqrt{\dfrac{1-\sin A}{1-\sin A}}
We know that a.b=ab\sqrt{a}.\sqrt{b}=\sqrt{ab}. By using this in the above expression, we get,
E=(1−sin⁡A)(1+sin⁡A).(1−sin⁡A)(1−sin⁡A)E=\sqrt{\dfrac{\left( 1-\sin A \right)}{\left( 1+\sin A \right)}.\dfrac{\left( 1-\sin A \right)}{\left( 1-\sin A \right)}}
E=(1−sin⁡A)2(1+sin⁡A)(1−sin⁡A)E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{\left( 1+\sin A \right)\left( 1-\sin A \right)}}
We know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}. By using this in the above expression, we get,
E=(1−sin⁡A)2(1)2−(sin⁡A)2E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sin A \right)}^{2}}}}
E=(1−sin⁡A)21−sin⁡2AE=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}
We know that cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 or 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta . By using this in the above expression, we get,
E=(1−sin⁡A)2cos⁡2AE=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{{{\cos }^{2}}A}}
We know that (a2b2)=(ab)2\left( \dfrac{{{a}^{2}}}{{{b}^{2}}} \right)={{\left( \dfrac{a}{b} \right)}^{2}}. By using this in the above expression, we get,
E=(1−sin⁡Acos⁡A)2E=\sqrt{{{\left( \dfrac{1-\sin A}{\cos A} \right)}^{2}}}
We know that x2=x\sqrt{{{x}^{2}}}=x. By using this in the above expression, we get,
E=1−sin⁡Acos⁡AE=\dfrac{1-\sin A}{\cos A}
We can also write the above expression as,
E=1cos⁡A−sin⁡Acos⁡AE=\dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}
We know that 1cos⁡θ=sec⁡θ\dfrac{1}{\cos \theta }=\sec \theta and sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta . By using this in the above expression, we get,
E = sec A – tan A
E = RHS
So, we get, LHS = RHS
Hence, we have proved that 1−sin⁡A1+sin⁡A=sec⁡A−tan⁡A\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A.

Note: In these type of questions containing the square root, some students try to convert the given angles into half angles but that is reliable in case of cos⁡θ\cos \theta but in case of sin⁡θ\sin \theta we should try to remove the square root by multiplying the conjugate of the denominator that is if the denominator is a + b, we will multiply (a – b) and vice versa. Then use trigonometric formulas to further simplify it.