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Question: Prove that: \(\dfrac{{\sin 2A}}{{\sin A}} - \dfrac{{\cos 2A}}{{\cos A}} = \sec A\)....

Prove that: sin⁡2Asin⁡A−cos⁡2Acos⁡A=sec⁡A\dfrac{{\sin 2A}}{{\sin A}} - \dfrac{{\cos 2A}}{{\cos A}} = \sec A.

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as the double angle formula of tangent, cos⁡(2x)=2cos⁡2x−1\cos \left( {2x} \right) = 2{\cos ^2}x - 1 and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x.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 answer:
In the given problem, we have to prove a trigonometric equality. 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. =sin⁡2Asin⁡A−cos⁡2Acos⁡A = \dfrac{{\sin 2A}}{{\sin A}} - \dfrac{{\cos 2A}}{{\cos A}}
We know the double angle formula for sine as sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x. Hence, we get,
2sin⁡Acos⁡Asin⁡A−cos⁡2Acos⁡A\dfrac{{2\sin A\cos A}}{{\sin A}} - \dfrac{{\cos 2A}}{{\cos A}}
Cancelling common factors in numerator and denominator, we get,
2cos⁡A−cos⁡2Acos⁡A2\cos A - \dfrac{{\cos 2A}}{{\cos A}}

Now, taking LCM of the expressions, we get,
2cos⁡2A−cos⁡2Acos⁡A\dfrac{{2{{\cos }^2}A - \cos 2A}}{{\cos A}}
Now, we know the double angle formula for cosine as cos⁡(2x)=2cos⁡2x−1\cos \left( {2x} \right) = 2{\cos ^2}x - 1. So, substituting this, we get,
2cos⁡2A−(2cos⁡2A−1)cos⁡A\dfrac{{2{{\cos }^2}A - \left( {2{{\cos }^2}A - 1} \right)}}{{\cos A}}
Opening the brackets and simplifying the expression,
2cos⁡2A−2cos⁡2A+1cos⁡A\dfrac{{2{{\cos }^2}A - 2{{\cos }^2}A + 1}}{{\cos A}}

Cancelling like terms with opposite signs, we get,
1cos⁡A\dfrac{1}{{\cos A}}
Now, we know that secant and cosine functions are the reciprocals of each other. So, we get,
sec⁡A\sec A
Now, R.H.S =sec⁡A = \sec A
As the left side of the equation is equal to the right side of the equation, we have,
sin⁡2Asin⁡A−cos⁡2Acos⁡A=sec⁡A\dfrac{{\sin 2A}}{{\sin A}} - \dfrac{{\cos 2A}}{{\cos A}} = \sec A
Hence, Proved.

Note: Besides these simple trigonometric formulae used in the problem, 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. We must remember the compound angle formula and double angle formula as they are widely used in many problems. One must take care while handling the calculations.