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

Prove: sin⁡A+cos⁡Acos⁡A−sin⁡A=tan⁡2A+sec⁡2A\dfrac{{\sin A + \cos A}}{{\cos A - \sin A}} = \tan 2A + \sec 2A

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right) and cos⁡(2x)=(2cos⁡2(x)−1)\cos \left( {2x} \right) = \left( {2{{\cos }^2}\left( x \right) - 1} \right) .

Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step answer:
In the given problem, we have to simplify the division of two trigonometric expressions and prove the required result. For proving the desired result, we need to first rationalise the expression so as to apply the double angle formulae for sine and cosine.
So, multiplying the numerator and denominator by (sin⁡A+cos⁡A)\left( {\sin A + \cos A} \right) to rationalise the fraction and to apply the double angle formulae, we get,
L.H.S. =sin⁡A+cos⁡Acos⁡A−sin⁡A = \dfrac{{\sin A + \cos A}}{{\cos A - \sin A}}
=sin⁡A+cos⁡Acos⁡A−sin⁡A×cos⁡A+sin⁡Acos⁡A+sin⁡A= \dfrac{{\sin A + \cos A}}{{\cos A - \sin A}} \times \dfrac{{\cos A + \sin A}}{{\cos A + \sin A}}
Now, we use the algebraic identity for square of a binomial (a+b)2=(a2+2ab+b2){\left( {a + b} \right)^2} = \left( {{a^2} + 2ab + {b^2}} \right) to evaluate the square of sum of sine and cosine of angle A, weget,
=sin⁡2A+cos⁡2A+2sin⁡A.cos⁡Acos⁡2A−sin⁡2A= \dfrac{{{{\sin }^2}A + {{\cos }^2}A + 2\sin A.\cos A}}{{{{\cos }^2}A - {{\sin }^2}A}}
Now, using the trigonometric identity sin⁡2(x)+cos⁡2(x)=1{\sin ^2}\left( x \right) + {\cos ^2}\left( x \right) = 1 ,
=1+sin⁡2Acos⁡2A= \dfrac{{1 + \sin 2A}}{{\cos 2A}}
Distributing the denominator to both the terms in the numerator separately, we get,
=1cos⁡2A+sin⁡2Acos⁡2A= \dfrac{1}{{\cos 2A}} + \dfrac{{\sin 2A}}{{\cos 2A}}
Now, using the basic using some of the simple trigonometric formulae such as sec⁡(x)=1cos⁡(x)\sec (x) = \dfrac{1}{{\cos (x)}} and tan⁡(x)=sin⁡(x)cos⁡(x)\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}} , we get,
=sec⁡2A+tan⁡2A= \sec 2A + \tan 2A
Rearranging the terms so as to match the left hand side of the above equation to the right hand side of the required equation, we get,
=tan⁡2A+sec⁡2A=R.H.S. = \tan 2A + \sec 2A = R.H.S.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. 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.