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Question: Prove that \[\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \tan A + \cot A\]...

Prove that tan⁡A1−cot⁡A+cot⁡A1−tan⁡A=1+tan⁡A+cot⁡A\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \tan A + \cot A?

Explanation

Solution

Hint : Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step answer :
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality tan⁡A1−cot⁡A+cot⁡A1−tan⁡A=1+tan⁡A+cot⁡A\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \tan A + \cot A
Now we consider the LHS
=tan⁡A1−cot⁡A+cot⁡A1−tan⁡A= \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosecant is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. From the reciprocal of the trigonometry ratios. Now the inequality is written as
=tan⁡A1−1tan⁡A+1tan⁡A1−tan⁡A= \dfrac{{\tan A}}{{1 - \dfrac{1}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}}
No we take the LCM for the both terms and this is written as
=tan⁡Atan⁡A−1tan⁡A+1tan⁡A1−tan⁡A= \dfrac{{\tan A}}{{\dfrac{{\tan A - 1}}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}}
On simplifying we have
=tan⁡2Atan⁡A−1+1tan⁡A(1−tan⁡A)= \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} + \dfrac{1}{{\tan A(1 - \tan A)}}
Take a minus as a common and the second term is rewritten as
=tan⁡2Atan⁡A−1−1tan⁡A(tan⁡A−1)= \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} - \dfrac{1}{{\tan A(\tan A - 1)}}
Now again taking the LCM for the both the terms we have
=tan⁡3A−1tan⁡A(tan⁡A−1)= \dfrac{{{{\tan }^3}A - 1}}{{\tan A(\tan A - 1)}}
The above inequality, in the numerator it is in the form of a3−b3{a^3} - {b^3}, we have standard algebraic identity for this and it is given as a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) Using this identity the above inequality is written as
=(tan⁡A−1)(tan⁡2A+1+tan⁡A)tan⁡A(tan⁡A−1)= \dfrac{{(\tan A - 1)({{\tan }^2}A + 1 + \tan A)}}{{\tan A(\tan A - 1)}}
On cancelling the terms we have
=(tan⁡2A+1+tan⁡A)tan⁡A= \dfrac{{({{\tan }^2}A + 1 + \tan A)}}{{\tan A}}
Now take the denominator value to the each every term of the numerator so we have
=tan⁡2Atan⁡A+1tan⁡A+tan⁡Atan⁡A= \dfrac{{{{\tan }^2}A}}{{\tan A}} + \dfrac{1}{{\tan A}} + \dfrac{{\tan A}}{{\tan A}}
On simplifying we have
=1+tan⁡A+cot⁡A= 1 + \tan A + \cot A
=RHS= RHS
Here we have proved LHS = RHS.

Note : The question involves the trigonometric functions. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.