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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles4 Marks
Question
If $\tan\text{x}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},$ then show that $\sin\alpha+\cos\alpha=\sqrt{2}\cos\text{x}.$
✓

Answer

$\tan\text{x}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$
$\Rightarrow\tan\text{x}=\frac{\tan\alpha-1}{\tan\alpha+1}$ [Dividing both numerator and denominator by $\cos\alpha$]
$\Rightarrow\tan\theta=\frac{\tan\alpha-\tan\frac{\pi}{4}}{1+ \tan\frac{\pi}{4}.\tan\alpha}$
$\Rightarrow\tan\theta=\tan\Big(\alpha-\frac{\pi}{4}\Big)$
$\Rightarrow\theta=\alpha-\frac{\pi}{4}$ [Removing tan form both sides]
$\Rightarrow\cos\theta=\cos\Big(\alpha-\frac{\pi}{4}\Big)$ [Taking cos on both sides]
$\Rightarrow\cos\theta=\cos\alpha.\cos\frac{\pi}{4}+\sin\alpha.\sin\frac{\pi}{4}$
$\Rightarrow\cos\theta=\cos\alpha.\frac{1}{\sqrt{2}}+\sin\alpha.\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{\cos\alpha+\sin\alpha}{\sqrt{2}}$
$\Rightarrow\sqrt{2}\cos\theta=\sin\alpha+\cos\alpha$
Hence proved.

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