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Question: How do you prove \(\dfrac{{1 + \tan x}}{{1 + \cot x}}\)?...

How do you prove 1+tan⁡x1+cot⁡x\dfrac{{1 + \tan x}}{{1 + \cot x}}?

Explanation

Solution

In order to simplify the above statement, rewrite the expression using the rule of trigonometry that cot⁡x\cot xis equal to the reciprocal of tan⁡x\tan x,and after taking the LCM in the denominator you’ll see that the numerator and denominator are both same ,so cancelling out both result in 1 and only tan⁡x\tan xis left with the expression which is the required answer.

Complete step by step solution:
We are given a trigonometric function1+tan⁡x1+cot⁡x\dfrac{{1 + \tan x}}{{1 + \cot x}} which we have to simplify

Recall properties of trigonometry that cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}}.So, now replacing cot⁡x\cot xwith this in our function.
=1+tan⁡x1+1tan⁡x= \dfrac{{1 + \tan x}}{{1 + \dfrac{1}{{\tan x}}}}

Taking LCM in the denominator
=1+tan⁡xtan⁡x+1tan⁡x= \dfrac{{1 + \tan x}}{{\dfrac{{\tan x + 1}}{{\tan x}}}}

Now rewriting the expression
=1+tan⁡x1+tan⁡x.(tan⁡x)= \dfrac{{1 + \tan x}}{{1 + \tan x}}.(\tan x)

Cancelling out the numerator and denominator, it results into 11

=1.(tan⁡x) =tan⁡x=sin⁡xcos⁡x = 1.(\tan x) \\\ = \tan x = \dfrac{{\sin x}}{{\cos x}} \\\

Therefore, the implication of function1+tan⁡x1+cot⁡x\dfrac{{1 + \tan x}}{{1 + \cot x}}is equal to tan⁡x or sin⁡xcos⁡x\tan x\,or\,\dfrac{{\sin x}}{{\cos x}}.

Formula:
tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}
cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}}

Note:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

2. Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.

If T is the smallest positive real number such that f(x+T)=f(x)f(x + T) = f(x) for all x, then T is called the fundamental period of f(x)f(x) .
Since sin⁡ (2nπ+θ)=sin⁡θ\sin \,(2n\pi + \theta ) = \sin \theta for all values of θ\theta and n∈\inN.

3. Even Function – A function f(x)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x)for all x in its domain.

Odd Function – A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x)for all x in its domain.

We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta