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

Prove that: cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=sin⁡A+cos⁡A\dfrac{{\cos A}}{{1 - \tan A}} + \dfrac{{\sin A}}{{1 - \cot A}} = \sin A + \cos A.

Explanation

Solution

The given question is related to trigonometric identities. We are given an equation and asked to prove that the left hand side of the equation is equal to the right hand side of the equation. To prove that, we will firstly take the left hand side of the equation and try to convert it to the right hand side of the equation using different trigonometric identities.
Formula used:
(a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)

Complete step-by-step solution:
To prove: cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=sin⁡A+cos⁡A\dfrac{{\cos A}}{{1 - \tan A}} + \dfrac{{\sin A}}{{1 - \cot A}} = \sin A + \cos A
Proof: At first, we will consider the left hand side of the equation,
L.H.S= cos⁡A1−tan⁡A+sin⁡A1−cot⁡A\dfrac{{\cos A}}{{1 - \tan A}} + \dfrac{{\sin A}}{{1 - \cot A}}
Expressing the numerator and denominator in terms ofsin⁡\sin andcos⁡\cos ,we get,
=cos⁡A1−sin⁡Acos⁡A+sin⁡A1−cos⁡Asin⁡A\dfrac{{\cos A}}{{1 - \dfrac{{\sin A}}{{\cos A}}}} + \dfrac{{\sin A}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}
Now, we will simplify for the denominator part,
=cos⁡Acos⁡A−sin⁡Acos⁡A+sin⁡Asin⁡A−cos⁡Asin⁡A\dfrac{{\cos A}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}} + \dfrac{{\sin A}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}
If we apply the fraction ruleabc=a×cb\dfrac{{\dfrac{a}{b}}}{c} = \dfrac{{a \times c}}{b}, we get,
=cos⁡A×cos⁡Acos⁡A−sin⁡A+sin⁡A×sin⁡Asin⁡A−cos⁡A\dfrac{{\cos A \times \cos A}}{{\cos A - \sin A}} + \dfrac{{\sin A \times \sin A}}{{\sin A - \cos A}}
Or, we can write the above equation as,
=cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡A\dfrac{{{{\cos }^2}A}}{{\cos A - \sin A}} + \dfrac{{{{\sin }^2}A}}{{\sin A - \cos A}}
Using the LCM method to add the above two fractions, we get,
=\dfrac{{{{\cos }^2}A - {{\sin }^2}A}}{{\cos A - \sin A}}$$$$$$ Using the identity,\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right),weget,=, we get, =\dfrac{{\left( {\cos A - \sin A} \right)\left( {\cos A + \sin A} \right)}}{{\cos A - \sin A}}Now,wecancancelthecommonfactor Now, we can cancel the common factor\cos A - \sin Afromthenumeratoranddenominator,weget,=from the numerator and denominator, we get, =\cos A + \sin A= =\sin A + \cos A$$, which is equal to the right hand side of the equation.
Hence, Left hand side=Right hand side.

Note: Alternatively, we can start the proof by taking the right hand side also. But that case might not be this much easy. It is advisable to start from the left hand side while solving such types of questions. It is very much important that the basic concepts of trigonometry must be clear. Like the relation of sin, cosine of an angle with tangent of an angle.