Question

alpha is divided into 2 parts such that the ratio of the tangents of the parts is k.if x is the difference between the two parts prove that (k+1)×sin(alpha)/(k-1)=sinx

Answer 1

Alpha=A/B
A-B=x. ; A=x+B
given: Tan A/ Tan B =k
Sin A/cos A*cos B/sin A=k
Sin A Cos B/Cos A sin B=k
Applying compendo dividendo rule
Sin A cos B+ cos A sinB/ sin A cos B-cosA sin B =k+1/k-1
Sin (A+B)/ sin (A-B)= k+1/k-1
doing reciprocal
Sin ( A-B)/sin (A+B)=k-1/k+1
Sin X. =k-1/k+1 sin alpha. ( since A=X+B)
W

Answer 2

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

Let, θ and β be the two parts of the angle α

Therefore, α = θ + β.

By question, x = θ - β. (assuming θ > β)

$and\thinspace \frac{\tan\theta}{\tan\beta} = k\\\\\implies \frac{\sin\theta\cdot \cos\beta}{\sin\beta\cdot \cos\theta} = \frac{k}{1}\\\\\text{Now, applying the rule of componendo and dividendo.} \\\\\implies \frac{\sin\theta\cdot \cos\beta + \sin\beta\cdot \cos\theta}{\sin\theta\cdot \cos\beta-\sin\beta \cdot \cos\theta}=\frac{(k+1)}{(k-1)}\\\\\implies \frac{\sin (\theta + \beta)}{\sin (\theta- \beta)} = \frac{(k + 1)}{(k - 1)}$

Since we know that α = θ + β. and  x = θ - β

$\implies \frac{\sin\alpha}{\sin x}=\frac{k+1}{k-1}\\\\\implies\bf\sin x=\frac{(k-1)}{k+1)}\times \sin\alpha$

Hence Proved.