Question

Hey !! ❤️

Question :-

If , α ≠ β and a tanα + b tanβ = ( a + b ) tan{( α + β ) / 2}

Then prove that ,

cosα / cosβ = a / b


Thanks!​

Answer 1

$\underline{\text{Proof :}}$

$\text{Given,}$

$\mathrm{a\:tan\alpha+b\:tan\beta=(a+b)\:tan\frac{\alpha+\beta}{2}}$

$\to \mathrm{a\frac{sin\alpha}{cos\alpha}+b\frac{sin\beta}{cos\beta}=a\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}}+b\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}}}$

$\to \small{\mathrm{a(\frac{sin\alpha}{cos\alpha}-\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}})+b (\frac{sin\beta}{cos\beta}-\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}})=0}}$

$\to \tiny{\mathrm{a\frac{sin\alpha\:cos\frac{\alpha+\beta}{2}-cos\alpha\:sin\frac{\alpha+\beta}{2}}{cos\alpha\:cos\frac{\alpha+\beta}{2}}+b\frac{sin\beta\:cos\frac{\alpha+\beta}{2}-cos\beta\:sin\frac{\alpha+\beta}{2}}{cos\beta\:cos\frac{\alpha+\beta}{2}}=0}}$

$\to \mathrm{a\frac{sin(\alpha-\frac{\alpha+\beta}{2})}{cos\alpha}+b\frac{sin(\beta-\frac{\alpha+\beta}{2})}{cos\beta}=0}$

$\to \mathrm{a\frac{sin\frac{\alpha-\beta}{2}}{cos\alpha}+b\frac{sin\frac{\beta-\alpha}{2}}{cos\beta}=0}$

$\to \mathrm{a\frac{sin\frac{\alpha-\beta}{2}}{cos\alpha}-b\frac{sin\frac{\alpha-\beta}{2}}{cos\beta}=0}$

$\to \mathrm{\frac{a}{cos\alpha}-\frac{b}{cos\beta}=0}$

$\to \mathrm{\frac{a}{cos\alpha}=\frac{b}{cos\beta}}$

$\to \boxed{\mathrm{\frac{cos\alpha}{cos\beta}=\frac{a}{b}}}$

$\text{Hence, proved.}$

Answer 2

Solution :

$\mathsf{atan\alpha+btan\beta=(a+b)tan(\dfrac{\alpha+\beta}{2})}\\\\\implies \mathsf{a\dfrac{sin\alpha}{cos\alpha}+b\dfrac{sin\alpha}{cos\alpha}=a\dfrac{sin\dfrac{\alpha+\beta}{2}}{cos\dfrac{\alpha+\beta}{2}}+b\dfrac{sin\dfrac{\alpha+\beta}{2}}{cos\dfrac{\alpha+\beta}{2}}}\\\\\implies \small{\mathsf{a(\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\dfrac{\alpha+\beta}{2}}{cos\dfrac{\alpha+\beta}{2}})+b (\dfrac{sin\beta}{cos\beta}-\dfrac{sin\dfrac{\alpha+\beta}{2}}{cos\dfrac{\alpha+\beta}{2}})=0}}$

$\implies \mathsf{a\dfrac{sin\alpha\:cos\dfrac{\alpha+\beta}{2}-cos\alpha\:sin\dfrac{\alpha+\beta}{2}}{cos\alpha\:cos\dfrac{\alpha+\beta}{2}}+b\dfrac{sin\beta\:cos\dfrac{\alpha+\beta}{2}-cos\beta\:sin\dfrac{\alpha+\beta}{2}}{cos\beta\:cos\dfrac{\alpha+\beta}{2}}=0}$

$\implies \mathsf{\dfrac{1}{cos\dfrac{\alpha+\beta}{2}}(a\dfrac{sin\alpha\:cos\dfrac{\alpha+\beta}{2}-cos\alpha\:sin\dfrac{\alpha+\beta}{2}}{cos\alpha}+b\dfrac{sin\beta\:cos\dfrac{\alpha+\beta}{2}-cos\beta\:sin\dfrac{\alpha+\beta}{2}}{cos\beta})=0}\\\\\implies \mathsf{a\dfrac{sin(\alpha-\dfrac{\alpha+\beta}{2})}{cos\alpha}+b\dfrac{sin(\beta-\dfrac{\alpha+\beta}{2})}{cos\beta}=0}$

$\implies \mathsf{a\dfrac{sin(\dfrac{2\alpha-\alpha-\beta}{2})}{cos\alpha}+b\dfrac{sin(\dfrac{2\beta-\alpha-\beta}{2})}{cos\beta}=0}\\\\\implies \mathsf{a\dfrac{sin(\dfrac{\alpha-\beta}{2})}{cos\alpha}-b\dfrac{sin(\dfrac{\alpha-\beta}{2})}{cos\beta}=0}\\\\\implies \mathsf{sin(\dfrac{\alpha-\beta}{2})(\dfrac{a}{cos\alpha}-\dfrac{b}{cos\beta})=0}$

$\implies \mathsf{\dfrac{a}{cos\alpha}-\dfrac{b}{cos\beta}=0}\\\\\implies \mathsf{\dfrac{a}{cos\alpha}=\dfrac{b}{cos\beta}}\\\\\implies \mathsf{\dfrac{cos\alpha}{cos\beta}=\dfrac{a}{b}}$

Step-by-step explanation:

We start by using the formula :

tan A = sin A / cos A .

Then transpose the values and go on taking commons until we are proving the statement .

Another formula used in the above proof is :

sin ( X - Y ) = sin X cos Y - cos X sin Y .