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Question: How do you prove \(\dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\sin A}}{{1 - \cot A}}\)?...

How do you prove cos⁡ A1−tan⁡A=sin⁡A1−cot⁡A\dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\sin A}}{{1 - \cot A}}?

Explanation

Solution

Here we have to solve the trigonometric functions. We have to prove left side of the equation to the right side of the equation. Firstly, we will solve left side of the equation by converting tan⁡θ\tan \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta by using the formula tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and then we will use different trigonometric identities to solve the equation such as cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 and similarly we will solve right side of the equation by converting cot⁡θ\cot \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta by using the formula cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }} and then we will use different trigonometric identities.

Complete step by step answer:
We have cos⁡ A1−tan⁡A=sin⁡A1−cot⁡A\dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\sin A}}{{1 - \cot A}}
Consider left side of the equation i.e., cos⁡ A1−tan⁡A\dfrac{{\cos \,A}}{{1 - \tan A}}
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}. So,
⇒cos⁡ A1−tan⁡A=cos⁡ A1−sin⁡Acos⁡A\Rightarrow \dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\cos \,A}}{{1 - \dfrac{{\sin A}}{{\cos A}}}}
⇒cos⁡ Acos⁡A−sin⁡Acos⁡A\Rightarrow \dfrac{{\cos \,A}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}
Simplifying the above equation. We get,
⇒cos⁡2 Acos⁡A−sin⁡A\Rightarrow \dfrac{{{{\cos }^2}\,A}}{{\cos A - \sin A}}

Multiply and divide the above equation by the conjugate of cos⁡A−sin⁡A\cos A - \sin A i.e., cos⁡A+sin⁡A\cos A + \sin A
⇒cos⁡2 Acos⁡A−sin⁡A×cos⁡A+sin⁡Acos⁡A+sin⁡A\Rightarrow \dfrac{{{{\cos }^2}\,A}}{{\cos A - \sin A}} \times \dfrac{{\cos A + \sin A}}{{\cos A + \sin A}}
Now using the identity (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}. We get,
⇒cos⁡2 A×(cos⁡A+sin⁡A)cos⁡2A−sin⁡2A\Rightarrow \dfrac{{{{\cos }^2}\,A \times (\cos A + \sin A)}}{{{{\cos }^2}A - {{\sin }^2}A}}
Now substituting cos⁡2A−sin⁡2A=cos⁡2A{\cos ^2}A - {\sin ^2}A = \cos 2A. We get,
⇒cos⁡2 A×(cos⁡A+sin⁡A)cos⁡2A\Rightarrow \dfrac{{{{\cos }^2}\,A \times (\cos A + \sin A)}}{{\cos 2A}}
Multiply and divide the above equation by 22. We get,
⇒2cos⁡2 A×(cos⁡A+sin⁡A)2cos⁡2A\Rightarrow \dfrac{{2{{\cos }^2}\,A \times (\cos A + \sin A)}}{{2\cos 2A}}

Now substituting 2cos⁡2A=cos⁡2A−12{\cos ^2}A = \cos 2A - 1 using the identity cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2{\cos ^2}\theta - 1. We get,
⇒(cos⁡2A−1)(cos⁡A+sin⁡A)2cos⁡2A\Rightarrow \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}
Therefore, cos⁡ A1−tan⁡A\dfrac{{\cos \,A}}{{1 - \tan A}} =(cos⁡2A−1)(cos⁡A+sin⁡A)2cos⁡2A = \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}
Now, consider right side of the equation i.e., sin⁡A1−cot⁡A\dfrac{{\sin A}}{{1 - \cot A}}
We know that  cot⁡θ=cos⁡θsin⁡θ\,\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}. So,
⇒sin⁡ A1−cot⁡A=sin⁡ A1−cos⁡Asin⁡A\Rightarrow \dfrac{{\sin \,A}}{{1 - \cot A}} = \dfrac{{\sin \,A}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}
⇒sin⁡ Asin⁡A−cos⁡Asin⁡A\Rightarrow \dfrac{{\sin \,A}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}
Simplifying the above equation. We get,
⇒sin⁡2 Asin⁡A−cos⁡A\Rightarrow \dfrac{{{{\sin }^2}\,A}}{{\sin A - \cos A}}

Taking negative signs common from the denominator. We get,
⇒−sin⁡2 Acos⁡A−sin⁡A\Rightarrow \dfrac{{ - {{\sin }^2}\,A}}{{\cos A - \sin A}}
Multiply and divide the above equation by the conjugate of cos⁡A−sin⁡A\cos A - \sin A i.e., cos⁡A+sin⁡A\cos A + \sin A
⇒−sin⁡2 Acos⁡A−sin⁡A×cos⁡A+sin⁡Acos⁡A+sin⁡A\Rightarrow \dfrac{{ - {{\sin }^2}\,A}}{{\cos A - \sin A}} \times \dfrac{{\cos A + \sin A}}{{\cos A + \sin A}}
Now using the identity (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}. We get,
⇒−sin⁡2 A(cos⁡A+sin⁡A)cos⁡2A−sin⁡2A\Rightarrow \dfrac{{ - {{\sin }^2}\,A(\cos A + \sin A)}}{{{{\cos }^2}A - {{\sin }^2}A}}
Now substituting cos⁡2A−sin⁡2A=cos⁡2A{\cos ^2}A - {\sin ^2}A = \cos 2A. We get,
⇒−sin⁡2 A×(cos⁡A+sin⁡A)cos⁡2A\Rightarrow \dfrac{{ - {{\sin }^2}\,A \times (\cos A + \sin A)}}{{\cos 2A}}

Multiply and divide the above equation by 22. We get,
⇒−2sin⁡2 A×(cos⁡A+sin⁡A)2cos⁡2A\Rightarrow \dfrac{{ - 2{{\sin }^2}\,A \times (\cos A + \sin A)}}{{2\cos 2A}}
Now substituting 2sin⁡2A=1−cos⁡2A2{\sin ^2}A = 1 - \cos 2A using the identitycos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta . We get
⇒−(1−cos⁡2A)×(cos⁡A+sin⁡A)2cos⁡2A\Rightarrow \dfrac{{ - (1 - \cos 2A) \times (\cos A + \sin A)}}{{2\cos 2A}}
Simplifying the above equation. We get,
⇒(cos⁡2A−1)×(cos⁡A+sin⁡A)2cos⁡2A\Rightarrow \dfrac{{(\cos 2A - 1) \times (\cos A + \sin A)}}{{2\cos 2A}}
Therefore, sin⁡A1−cot⁡A\dfrac{{\sin A}}{{1 - \cot A}} =(cos⁡2A−1)(cos⁡A+sin⁡A)2cos⁡2A = \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}

Since the left side of the equation is equal to the right side of the equation.Hence proved.

Note: In these types of problems in which we have to prove left side of the equation is equal to the right side of the equation if we are unable to prove one side equal to the other side we can solve both side of the equation and prove their results equal to one another. In these types of problems use trigonometric identity carefully and carefully identify by what number or equations we can multiply the equation. Note that if we are multiplying something in the numerator we also multiply in the denominator.