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

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

Explanation

Solution

Hint:First of all, substitute tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A} and cot⁡A=cos⁡Asin⁡A\cot A=\dfrac{\cos A}{\sin A} in LHS of the given expression. Now, simplify the expression and use the identity a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to get the desired result.

Complete step-by-step answer:
In this question, we have to prove that
cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡A+sin⁡A\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A
Let us consider the LHS of the equation given in the question, we get,
E=cos⁡A1−tan⁡A+sin⁡A1−cot⁡AE=\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }. By using this in the above expression, we get,
E=cos⁡A1−sin⁡Acos⁡A+sin⁡A1−cos⁡Asin⁡AE=\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}+\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}}
By simplifying the above equation, we get,
E=cos⁡Acos⁡A−sin⁡Acos⁡A+sin⁡Asin⁡A−cos⁡Asin⁡AE=\dfrac{\cos A}{\dfrac{\cos A-\sin A}{\cos A}}+\dfrac{\sin A}{\dfrac{\sin A-\cos A}{\sin A}}
E=(cos⁡A).(cos⁡A)(cos⁡A−sin⁡A)+(sin⁡A).(sin⁡A)(sin⁡A−cos⁡A)E=\dfrac{\left( \cos A \right).\left( \cos A \right)}{\left( \cos A-\sin A \right)}+\dfrac{\left( \sin A \right).\left( \sin A \right)}{\left( \sin A-\cos A \right)}
We can also write the above equation as,
E=cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡AE=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}+\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}
By taking (– 1) common from the second term of the above expression, we get,
E=cos⁡2A(cos⁡A−sin⁡A)+sin⁡2A−1(cos⁡A−sin⁡A)E=\dfrac{{{\cos }^{2}}A}{\left( \cos A-\sin A \right)}+\dfrac{{{\sin }^{2}}A}{-1\left( \cos A-\sin A \right)}
We can also write the above expression as,
E=cos⁡2A(cos⁡A−sin⁡A)−sin⁡2A(cos⁡A−sin⁡A)E=\dfrac{{{\cos }^{2}}A}{\left( \cos A-\sin A \right)}-\dfrac{{{\sin }^{2}}A}{\left( \cos A-\sin A \right)}
By taking (cos A – sin A) as LCM in the above expression, we get,
E=(cos⁡2A−sin⁡2A)(cos⁡A−sin⁡A)E=\dfrac{\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)}{\left( \cos A-\sin A \right)}
We know that a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right). By substituting a = cos A and b = sin A, we get,
(cos⁡2A−sin⁡2A)=(cos⁡A+sin⁡A)(cos⁡A−sin⁡A)\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)=\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)
By using this in the above expression, we get,
E=(cos⁡A+sin⁡A)(cos⁡A−sin⁡A)(cos⁡A−sin⁡A)E=\dfrac{\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)}{\left( \cos A-\sin A \right)}
Now, by canceling out the like terms in the above expression, we get, E=cos⁡A+sin⁡AE=\cos A+\sin A which is equal to RHS.
So, LHS = RHS.
Hence, we have proved that cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡A+sin⁡A\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A

Note: In this question, students must note that it is always better to convert the expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . Also, students must take care while taking (– 1) common in the above solution. We have done that step to make the denominator of both the fractions similar. Also, in these questions involving various trigonometric ratios, it is advisable to cross-check every step.Students should remember Trigonometric ratios and identities for solving these types of problems.