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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
If $\cos(\theta+\phi)=\text{m}\cos(\theta-\phi),$ then prove that $\tan \theta=\frac{1-\text{m}}{1+\text{m}}\cot\phi.$
[Hint: Express $\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{\text{m}}{1}$ and apply Componendo and Dividendo]
✓

Answer

Given that: $\cos(\theta+\phi)=\text{m}\cos(\theta-\phi)$
$\Rightarrow\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{\text{m}}{1}$
Using componendo and dividend rule, we get
$\frac{\cos(\theta+\phi)+\cos(\theta-\phi)}{\cos(\theta+\phi)-\cos(\theta-\phi)}=\frac{\text{m}+1}{\text{m}-1}$
$\Big[\cos\text{C}-\cos\text{D}=-2\sin\Big(\frac{\text{C+D}}{2}\Big)\sin\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$\Big[\cos\text{C}+\cos\text{D}=2\cos\Big(\frac{\text{C+D}}{2}\Big)\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$
$\Rightarrow\frac{2\cos\Big(\frac{\theta+\phi+\theta-\phi}{2}\Big).\cos\Big(\frac{\theta+\phi-\theta+\phi}{2}\Big)}{-2\sin\Big(\frac{\theta+\phi+\theta-\phi}{2}\Big).\sin\Big(\frac{\theta+\phi-\theta+\phi}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\frac{\cos\theta\cos\phi}{-\sin\theta\sin\phi}=\frac{\text{m}+1}{\text{m}-1}\Rightarrow-\cot\theta.\cot\phi=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\frac{-\cot\phi}{\tan\theta}=\frac{\text{m}+1}{\text{m}-1}=-\frac{1+\text{m}}{1-\text{m}}$
$\Rightarrow\tan\theta(1+\text{m})=(1-\text{m})\cot\phi$
$\Rightarrow\tan\theta=\frac{1-\text{m}}{1+\text{m}}\cot\phi.$ Hence proved.

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