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Question: Prove \[\dfrac{{\cos A}}{{\sin B}} - \dfrac{{\sin A}}{{\cos B}} = \dfrac{{2\cos \left( {A + B} \righ...

Prove cos⁡Asin⁡B−sin⁡Acos⁡B=2cos⁡(A+B)sin⁡2B\dfrac{{\cos A}}{{\sin B}} - \dfrac{{\sin A}}{{\cos B}} = \dfrac{{2\cos \left( {A + B} \right)}}{{\sin 2B}}

Explanation

Solution

To solve this question first we chose any side of the equation. So, first, we take the left-hand side then we take LCM in the denominator, and then use the formula of the sum of angles with cos trigonometric function and then we multiply and divide by 2 to change the denominator part in double angle and then we get the final answer which is equal to the right-hand side of the equation.

Complete step by step solution:
We have to prove cos⁡Asin⁡B−sin⁡Acos⁡B=2cos⁡(A+B)sin⁡2B\dfrac{{\cos A}}{{\sin B}} - \dfrac{{\sin A}}{{\cos B}} = \dfrac{{2\cos \left( {A + B} \right)}}{{\sin 2B}}
Formula used in the questions are:
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B and 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta these are the two formulas which are used. Here the first formula is the sum of angles and double angle of the trigonometric function.
We will take the left-hand side part and prove the right-hand side.
Left-hand side is
⇒cos⁡Asin⁡B−sin⁡Acos⁡B\Rightarrow \dfrac{{\cos A}}{{\sin B}} - \dfrac{{\sin A}}{{\cos B}}
Now taking the LCM n denominator
⇒cos⁡Acos⁡B−sin⁡Asin⁡Bsin⁡Bcos⁡B\Rightarrow \dfrac{{\cos A\cos B - \sin A\sin B}}{{\sin B\cos B}}
Now using the formula of sum of two angle of cos trigonometric function
cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B
⇒cos⁡(A+B)sin⁡Bcos⁡B\Rightarrow \dfrac{{\cos \left( {A + B} \right)}}{{\sin B\cos B}}
Now we are going to multiply and divide by 2
⇒2cos⁡(A+B)2sin⁡Bcos⁡B\Rightarrow \dfrac{{2\cos \left( {A + B} \right)}}{{2\sin B\cos B}}
Now using the formula of double angle of sin trigonometric function
2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta
⇒2cos⁡(A+B)sin⁡2B\Rightarrow \dfrac{{2\cos \left( {A + B} \right)}}{{\sin 2B}}
The obtained answer is equal to the right-hand side of the question.
Hence proved.

Note:
Although this question is easy. But in order to solve this question, you must know all the formulas of trigonometry and students often make mistakes in using the formulas and the calculations. This can be done from the right-hand side, at last, we get an answer as the left-hand side of the question. Students must know the formulas of double angle, the sum of two angles, triple angle, half-angle formulas of all the trigonometric functions.