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Question: Prove that \(\cot A - \tan A = 2\cot 2A\)....

Prove that cot⁡A−tan⁡A=2cot⁡2A\cot A - \tan A = 2\cot 2A.

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as the value of tangent and cotangent functions in terms of sine and cosine. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us.

Complete step-by-step solution:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications.
For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. =cot⁡A−tan⁡A = \cot A - \tan A
So, we know the values of trigonometric functions tangent and cotangent as sin⁡xcos⁡x\dfrac{{\sin x}}{{\cos x}} and cos⁡xsin⁡x\dfrac{{\cos x}}{{\sin x}}.
So, we simplify the left hand side of the equation using these formulae as,
=cos⁡Asin⁡A−sin⁡Acos⁡A= \dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\cos A}}
Now, we take LCM of the rational trigonometric expressions to simplify the expressions. So, we get,
=cos⁡Acos⁡Asin⁡Acos⁡A−sin⁡Asin⁡Asin⁡Acos⁡A= \dfrac{{\cos A\cos A}}{{\sin A\cos A}} - \dfrac{{\sin A\sin A}}{{\sin A\cos A}}
Now, since the denominators of both the rational expressions is same, so combining the numerators, we get,
=cos⁡2A−sin⁡2Asin⁡Acos⁡A= \dfrac{{{{\cos }^2}A - {{\sin }^2}A}}{{\sin A\cos A}}
Now, we have to apply the double angle formula of cosine cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x in the numerator. So, we get,
=cos⁡2Asin⁡Acos⁡A= \dfrac{{\cos 2A}}{{\sin A\cos A}}
Now, we multiply and divide the expression by two. So, we get,
=2cos⁡2A2sin⁡Acos⁡A= \dfrac{{2\cos 2A}}{{2\sin A\cos A}}
Now, we use the double angle formula of sine in the denominator sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x. So, we get,
=2cos⁡2Asin⁡2A= \dfrac{{2\cos 2A}}{{\sin 2A}}
Now, we also know that cotangent is the ratio of cosine and sine functions. So, we get,
=2cot⁡2A= 2\cot 2A
Also, R.H.S. =2cot⁡2A = 2\cot 2A
Since, L.H.S.=R.H.S.L.H.S. = R.H.S.. So, we get, cot⁡A−tan⁡A=2cot⁡2A\cot A - \tan A = 2\cot 2A
Hence, Proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. One must take care while handling the calculations.