Solveeit Logo
Login

Question

Question: Show that \(\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\cos A+\sin A\)....

Show 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.

Explanation

Solution

We have sum of two terms in the left-hand side of 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. We multiply cos⁡A\cos A with cos⁡A1−tan⁡A\dfrac{\cos A}{1-\tan A} and sin⁡A\sin A with sin⁡A1−cot⁡A\dfrac{\sin A}{1-\cot A} on both numerator and denominator. Then we add them as the denominators are the same. Then we use the identity of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to factor the numerator. We also need to mention the conditions.

Complete step by step answer:
We multiply cos⁡A\cos A to the numerator and denominator of cos⁡A1−tan⁡A\dfrac{\cos A}{1-\tan A}.
We know tan⁡A=sin⁡Acos⁡A\tan A=\dfrac{\sin A}{\cos A}. This gives tan⁡A.cos⁡A=sin⁡A\tan A.\cos A=\sin A.So,
cos⁡A1−tan⁡A×cos⁡Acos⁡A=cos⁡2Acos⁡A−sin⁡A\dfrac{\cos A}{1-\tan A}\times \dfrac{\cos A}{\cos A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}
We multiply sin⁡A\sin A to the numerator and denominator of sin⁡A1−cot⁡A\dfrac{\sin A}{1-\cot A}.
We know cot⁡A=cos⁡Asin⁡A\cot A=\dfrac{\cos A}{\sin A}. This gives cot⁡A.sin⁡A=cos⁡A\cot A.\sin A=\cos A.
So, sin⁡A1−cot⁡A×sin⁡Asin⁡A=sin⁡2Asin⁡A−cos⁡A\dfrac{\sin A}{1-\cot A}\times \dfrac{\sin A}{\sin A}=\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}.

The equation becomes cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡A\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}+\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}.
To make the denominators same we take a negative sign common.
cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡2Acos⁡A−sin⁡A−sin⁡2Acos⁡A−sin⁡A\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}-\dfrac{{{\sin }^{2}}A}{\cos A-\sin A}.
The addition gives cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=cos⁡2A−sin⁡2Acos⁡A−sin⁡A\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\cos A-\sin A}
The numerator is in the form of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right). We now apply the identity theorem for the term cos⁡2A−sin⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A. We assume the values a=cos⁡A,b=sin⁡Aa=\cos A,b=\sin A.

Applying the theorem, we get cos⁡2A−sin⁡2A=(cos⁡A+sin⁡A)(cos⁡A−sin⁡A){{\cos }^{2}}A-{{\sin }^{2}}A=\left( \cos A+\sin A \right)\left( \cos A-\sin A \right).
The equation becomes cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=(cos⁡A+sin⁡A)(cos⁡A−sin⁡A)(cos⁡A−sin⁡A)\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\dfrac{\left( \cos A+\sin A \right)\left( \cos A-\sin A \right)}{\left( \cos A-\sin A \right)}.
We can now eliminate the (cos⁡A−sin⁡A)\left( \cos A-\sin A \right) from both denominator and numerator.
The equation becomes 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.
Thus verified cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=(cos⁡A+sin⁡A)\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\left( \cos A+\sin A \right).

Note: It is important to remember that the condition to eliminate the (cos⁡A−sin⁡A)\left( \cos A-\sin A \right) from both denominator and numerator is (cos⁡A−sin⁡A)≠0\left( \cos A-\sin A \right)\ne 0. No domain is given for the variable AA. The value of tan⁡A≠0\tan A\ne 0 is essential. The simplified condition will be A≠nπ,n∈ZA\ne n\pi ,n\in \mathbb{Z}. We also have the multiple angle theorem of cos⁡2A−sin⁡2A=cos⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A.