Solveeit Logo
Login

Question

Question: Find the value of \({{\left( \dfrac{1+i\tan \theta }{1-i\tan \theta } \right)}^{n}}-\dfrac{1+i\tan n...

Find the value of (1+itan⁡θ1−itan⁡θ)n−1+itan⁡nθ1−itan⁡nθ{{\left( \dfrac{1+i\tan \theta }{1-i\tan \theta } \right)}^{n}}-\dfrac{1+i\tan n\theta }{1-i\tan n\theta }
(A) 0\left( A \right)\text{ 0}
(B) 1\left( B \right)\text{ 1}

Explanation

Solution

In this question we have been given a trigonometric expression for which we have to find the value. We will solve this question by expanding the given expression by writing tan⁡θ\tan \theta in the form of sin⁡θcos⁡θ\dfrac{\sin \theta }{\cos \theta } and then taking the lowest common multiple. We will then use De moivre’s theorem which is given as (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta and simplify the expression to get the required solution.

Complete step by step answer:
We have the expression given to us as:
⇒(1+itan⁡θ1−itan⁡θ)n−1+itan⁡nθ1−itan⁡nθ\Rightarrow {{\left( \dfrac{1+i\tan \theta }{1-i\tan \theta } \right)}^{n}}-\dfrac{1+i\tan n\theta }{1-i\tan n\theta }
Now we know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } therefore, on substituting it in the expression, we get:
⇒(1+isin⁡θcos⁡θ1−isin⁡θcos⁡θ)n−1+isin⁡nθcos⁡nθ1−isin⁡nθcos⁡nθ\Rightarrow {{\left( \dfrac{1+i\dfrac{\sin \theta }{\cos \theta }}{1-i\dfrac{\sin \theta }{\cos \theta }} \right)}^{n}}-\dfrac{1+i\dfrac{\sin n\theta }{\cos n\theta }}{1-i\dfrac{\sin n\theta }{\cos n\theta }}
On taking the lowest common multiple in the fractions, we get:
⇒(cos⁡θ+isin⁡θcos⁡θcos⁡θ−isin⁡θcos⁡θ)n−cos⁡nθ+isin⁡nθcos⁡nθcos⁡nθ−isin⁡nθcos⁡nθ\Rightarrow {{\left( \dfrac{\dfrac{\cos \theta +i\sin \theta }{\cos \theta }}{\dfrac{\cos \theta -i\sin \theta }{\cos \theta }} \right)}^{n}}-\dfrac{\dfrac{\cos n\theta +i\sin n\theta }{\cos n\theta }}{\dfrac{\cos n\theta -i\sin n\theta }{\cos n\theta }}
On cancelling the terms, we get:
⇒(cos⁡θ+isin⁡θcos⁡θ−isin⁡θ)n−cos⁡nθ+isin⁡nθcos⁡nθ−isin⁡nθ\Rightarrow {{\left( \dfrac{\cos \theta +i\sin \theta }{\cos \theta -i\sin \theta } \right)}^{n}}-\dfrac{\cos n\theta +i\sin n\theta }{\cos n\theta -i\sin n\theta }
Now we know from De moivre’s theorem that (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta therefore, on using in the above expression, we get:
⇒cos⁡nθ+isin⁡nθcos⁡nθ−isin⁡nθ−cos⁡nθ+isin⁡nθcos⁡nθ−isin⁡nθ\Rightarrow \dfrac{\cos n\theta +i\sin n\theta }{\cos n\theta -i\sin n\theta }-\dfrac{\cos n\theta +i\sin n\theta }{\cos n\theta -i\sin n\theta }
Now we can see that both the terms are the same therefore on simplifying, we get:
⇒0\Rightarrow 0, which is the required solution.

So, the correct answer is “Option A”.

Note: It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken.
The various trigonometric identities and formulae should be remembered while doing these types of sums. the various Pythagorean identities should also be remembered while doing these type of questions.
To simplify any given equation, it is good practice to convert all the identities into sin⁡\sin and cos⁡\cos for simplifying.
If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.