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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Prove that $2\tan^{-1}\bigg(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\bigg)=\cos^{-1}\Big(\frac{\text{a}\cos\theta+b}{\text{a}+\text{b}\cos\theta}\Big)$
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Answer

$\text{L.H.S}=2\tan^{-1}\bigg(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\bigg)$
$=\cos^{-1}\begin{Bmatrix}\frac{1-\Big(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\Big)^2}{1+\Big(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\Big)^2}\end{Bmatrix}$ $\Big[\because\ 2\tan^{-1}(\text{x})=\cos^{-1}\Big\{\frac{1-\text{x}^2}{1-\text{x}^2}\Big\}\Big]$
$=\cos^{-1}\Bigg\{\frac{1-\frac{\text{a}-\text{b}}{\text{a}++\text{b}}\tan^{2}\frac{\theta}{2}}{1+\frac{\text{a}-\text{b}}{\text{a}++\text{b}}\tan^{2}\frac{\theta}{2}}\Bigg\}$
$=\cos^{-1}\Bigg\{\frac{\text{a}+\text{b}-(\text{a}-\text{b})\tan^{2}\frac{\theta}{2}}{\text{a}+\text{b}+(\text{a}-\text{b})\tan^{2}\frac{\theta}{2}}\Bigg\}$
$=\cos^{-1}\Bigg\{\frac{\text{a}+\text{b}-\text{a}\tan^2\frac{\theta}{2}+\text{b}\tan^2\frac{\theta}{2}}{\text{a}+\text{b}+\text{a}\tan^2\frac{\theta}{2}-\text{b}\tan^2\frac{\theta}{2}}\Bigg\}$
$=\cos\begin{Bmatrix}\frac{\text{a}\Big(1-\tan^2\frac{\theta}{2}\Big)+\text{b}\Big(1+\tan^2\frac{\theta}{2}\Big)}{\text{a}\Big(1+\tan^2\frac{\theta}{2}\Big)+\text{b}\Big(1-\tan^2\frac{\theta}{2}\Big)}\end{Bmatrix}$
$=\cos^{-1}\begin{Bmatrix}\frac{\text{a}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}\Bigg(\frac{1+\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)}{\text{a}\Bigg(\frac{1+\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\Bigg)}\end{Bmatrix}$ $\Big[\text{Dividing N' and D' by }1+\tan^2\frac{\theta}{2}\Big]$
$=\cos^{-1}\begin{Bmatrix}\frac{\text{a}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}}{\text{a}+\text{b}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\Bigg)}\end{Bmatrix}$
$=\cos^{-1}\Big(\frac{\text{a}\cos\theta+b}{\text{a}+\text{b}\cos\theta}\Big)=\text{R.H.S}$

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