Question

If $a\tan \alpha + b\tan \beta = \left( {a + b} \right)\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ , where $\alpha \ne \beta $ , prove that $a\cos \beta = b\cos \alpha $

Answer 1

Hint: Start with the given expression $a\tan \alpha + b\tan \beta = \left( {a + b} \right)\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ and go step by step to transform it into the required form. Change the tangent function into the ratio of sine and cosine. Separate the terms with ‘a’ and ‘b’ and take common. Now use the sine subtraction formula, i.e. $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ . Cancel out the common terms on both sides. Simplify it further to obtain the required expression.

Complete step-by-step answer:
Here in the problem, we are given an expression $a\tan \alpha + b\tan \beta = \left( {a + b} \right)\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ . And with this given expression and using other trigonometric identities, we need to prove that $a\cos \beta = b\cos \alpha $ .
We can solve this by transforming the given expression into the form of the cosine ratio.
Let’s use the following property of tangent that defines a tangent function as the ratio of sine and cosine function, i.e. $\tan x = \dfrac{{\sin x}}{{\cos x}}$
$ \Rightarrow a\tan \alpha + b\tan \beta = \left( {a + b} \right)\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right) \Rightarrow a\dfrac{{\sin \alpha }}{{\cos \alpha }} + b\dfrac{{\sin \beta }}{{\cos \beta }} = \left( {a + b} \right)\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}$
Now we can remove the parenthesis and expand the expression, this will give:
$ \Rightarrow a\dfrac{{\sin \alpha }}{{\cos \alpha }} + b\dfrac{{\sin \beta }}{{\cos \beta }} = \left( {a + b} \right)\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}} \Rightarrow a\dfrac{{\sin \alpha }}{{\cos \alpha }} + b\dfrac{{\sin \beta }}{{\cos \beta }} = a\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}} + b\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}$
Let’s transpose the terms with ‘a’ and ‘b’ on separate sides in the equation
$ \Rightarrow a\dfrac{{\sin \alpha }}{{\cos \alpha }} - a\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}} = b\dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}} - b\dfrac{{\sin \beta }}{{\cos \beta }}$
Now combining the two fractions on both sides by taking ‘a’ and ‘b’ common in both sides
$ \Rightarrow a\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right) = - b\left( {\dfrac{{\sin \beta }}{{\cos \beta }} - \dfrac{{\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right)$
Let’s subtract the fractions and making them a single fraction but multiply the denominators together
$ \Rightarrow a\left( {\dfrac{{\sin \alpha \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right) - \sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \alpha }}{{\cos \alpha \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right) = - b\left( {\dfrac{{\sin \beta \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right) - \sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \beta }}{{\cos \beta \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right)$
As we know from the sine subtraction formula for two angles ‘a’ and ‘b’ :
$\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$
We can use the above identity in the previous expression in the numerator to simplify
$ \Rightarrow a\left( {\dfrac{{\sin \left( {\alpha - \dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \alpha \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right) = - b\left( {\dfrac{{\sin \left( {\beta - \dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \beta \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right)$
Now we know that $\alpha \ne \beta $ , so after multiplying with $\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ we get:
$ \Rightarrow a\left( {\dfrac{{\sin \left( {\alpha - \dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \alpha \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right) = - b\left( {\dfrac{{\sin \left( {\beta - \dfrac{{\alpha + \beta }}{2}} \right)}}{{\cos \beta \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}} \right) \Rightarrow a\dfrac{{\sin \left( {\dfrac{{2\alpha - \alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = - b\dfrac{{\sin \left( {\dfrac{{2\beta - \alpha - \beta }}{2}} \right)}}{{\cos \beta }}$
Let’s simplify the numerator and we get:
$ \Rightarrow a\dfrac{{\sin \left( {\dfrac{{2\alpha - \alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = - b\dfrac{{\sin \left( {\dfrac{{2\beta - \alpha - \beta }}{2}} \right)}}{{\cos \beta }} \Rightarrow a\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = - b\dfrac{{\sin \left( {\dfrac{{\beta - \alpha }}{2}} \right)}}{{\cos \beta }}$
We also know that for sine ratio, we can say: $ - \sin t = \sin \left( { - t} \right)$
Therefore, using that we get:
$ \Rightarrow a\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = - b\dfrac{{\sin \left( {\dfrac{{\beta - \alpha }}{2}} \right)}}{{\cos \beta }} \Rightarrow a\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = b\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \beta }}$
Now since we already know $\alpha \ne \beta $ , dividing both sides with $\sin \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ , we obtain:
$ \Rightarrow a\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \alpha }} = b\dfrac{{\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)}}{{\cos \beta }} \Rightarrow \dfrac{a}{{\cos \alpha }} = \dfrac{b}{{\cos \beta }}$
Now we can transpose the denominators to opposite sides, we get:
$ \Rightarrow \dfrac{a}{{\cos \alpha }} = \dfrac{b}{{\cos \beta }} \Rightarrow a\cos \beta = b\cos \alpha $
Therefore, we get the required expression $ \Rightarrow a\cos \beta = b\cos \alpha $ from the given expression in tangent.

Note: In questions like this, it is very clear that you need to transform the only given expression to change it into the required form since there is no other information given. Notice that the use of the sine formula i.e. $\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b$ was a very crucial part of the solution. Be careful with the signs of the trigonometric ratios and the angles.