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Question: Evaluate the value of \(\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}\) ?...

Evaluate the value of 1+sin⁡θ1−sin⁡θ\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} ?

Explanation

Solution

These types of problems are of the form fraction\sqrt{fraction} . So, the general solution of these types of problems is by rationalising the denominator by multiplying same expressions to the numerator and the denominator. Here too, we multiply 1+sin⁡θ\sqrt{1+\sin \theta } to the numerator and the denominator. After that, we apply some basic trigonometric formulae and arrive at our desired simplified form.

Complete step by step answer:
The given expression is
1+sin⁡θ1−sin⁡θ\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}
We start off the solution by multiplying 1+sin⁡θ\sqrt{1+\sin \theta } in the numerator and denominator both. The expression thus becomes,
⇒1+sin⁡θ1−sin⁡θ×1+sin⁡θ1+sin⁡θ\Rightarrow \sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}\times \sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}
The denominator becomes (1−sin⁡θ)2\sqrt{{{\left( 1-\sin \theta \right)}^{2}}} and the numerator becomes (1+sin⁡θ)(1−sin⁡θ)\sqrt{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)} . The expression thus becomes,
⇒(1+sin⁡θ)2(1−sin⁡θ)(1+sin⁡θ)\Rightarrow \dfrac{\sqrt{{{\left( 1+\sin \theta \right)}^{2}}}}{\sqrt{\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)}}
Which can be further simplified to
⇒(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\sqrt{\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)}}
Now, we know the formula of squares which states that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} . If we compare the denominator of the above expression with the formula of squares, we can say that a=1a=1 and b=sin⁡θb=\sin \theta . So, applying the formula in the above expression, the expression thus becomes,
⇒(1+sin⁡θ)(1−sin⁡2θ)\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\sqrt{\left( 1-{{\sin }^{2}}\theta \right)}}
We know that cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta . Applying this formula in the above expression, we get,
⇒(1+sin⁡θ)(cos⁡2θ)\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\sqrt{\left( {{\cos }^{2}}\theta \right)}}
This can be further simplified to
⇒(1+sin⁡θ)cos⁡θ\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\cos \theta }
We now split the above expression into two fractions. The expression thus becomes,
⇒1cos⁡θ+sin⁡θcos⁡θ\Rightarrow \dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }
We all know that the reciprocal trigonometric ratio sec⁡θ\sec \theta is equal to 1cos⁡θ\dfrac{1}{\cos \theta } . The above expression thus can be rewritten as,.
⇒sec⁡θ+sin⁡θcos⁡θ\Rightarrow \sec \theta +\dfrac{\sin \theta }{\cos \theta }
Also, we know that tan⁡θ\tan \theta is nothing but sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } . Thus, rewriting the above expression after applying this formula, we get,
⇒sec⁡θ+tan⁡θ\Rightarrow \sec \theta +\tan \theta
Therefore, we can conclude that the simplified form of the value of the given expression 1+sin⁡θ1−sin⁡θ\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }} is sec⁡θ+tan⁡θ\sec \theta +\tan \theta .

Note: While solving these types of problems, we should choose an appropriate term to rationalise the denominator. Also, we must be careful while simplifying the fractions, as students often miss out some terms and this leads to mistakes. We should express the given expression in the most simplified form possible.