If m =n ( +2 ), then prove that ( + ) = m+n m-n [Hint: Express ( +2 ) = m n and apply componendo and dividendo]

Given that: m =n ( +2 ) ( +2 ) = m n Using componendo and dividendo rule we get ( +2 )+ ( +2 )- = m+n m-n 2 ( +2 + 2 ). ( +2 - 2 ) 2 ( +2 + 2 ). ( +2 - 2 ) = m+n m-n + =2 A + B 2 . A-B 2 - =2 A + B 2 . A-B 2 ( + ). ( + ). = m+n m-n ( + ). = m+n m-n Hence proved.

If m =n ( +2 ), then prove that ( + ) = m+n m-n [Hint: Express ( …

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Question

If $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$ then prove that $\tan(\theta+\alpha)\cot\alpha=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$[Hint: Express $\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$ and apply componendo and dividendo]

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Answer

Given that: $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha)$
$\Rightarrow\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$
Using componendo and dividendo rule we get
$\Rightarrow\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\Rightarrow\frac{2\cos\big(\frac{\theta+2\alpha+\theta}{2}\big).\cos\big(\frac{\theta+2\alpha-\theta}{2}\big)}{2\sin\big(\frac{\theta+2\alpha+\theta}{2}\big).\sin\big(\frac{\theta+2\alpha-\theta}{2}\big)}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\begin{bmatrix}\because\sin\text{A}+\sin\text{B}=2\sin\frac{\text{A + B}}{2}.\cos\frac{\text{A}-\text{B}}{2}\\\sin\text{A}-\sin\text{B}=2\cos\frac{\text{A + B}}{2}.\sin\frac{\text{A}-\text{B}}{2}\end{bmatrix}$
$\Rightarrow\frac{\sin(\theta+\alpha).\cos\alpha}{\cos(\theta+\alpha).\sin\alpha}=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$
$\Rightarrow\tan(\theta+\alpha).\cot\alpha=\frac{\text{m}+\text{n}}{\text{m}-\text{n}}$ Hence proved.