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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
If $\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},$ then show that $\sin\alpha+\cos\alpha=\sqrt{2}\cos\theta.$
[Hint: Express $\tan\theta=\tan(\alpha-\frac{\pi}{4})\theta=\alpha-\frac{\pi}{4}$ ]
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Answer

Given that: $\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}$
$\Rightarrow\tan\theta=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}$
$\Rightarrow\tan\theta=\frac{\tan\alpha-1}{\tan\alpha+1}=\frac{\tan\alpha-\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{4}\tan\alpha}$$[\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}]$
$\Rightarrow\tan\theta=\tan\Big(\alpha-\frac{\pi}{4}\Big)$
$\therefore\theta=\alpha-\frac{\pi}{4}$
$\Rightarrow\cos\theta=\cos\Big(\alpha-\frac{\pi}{4}\Big)$
$\Rightarrow\cos\theta=\cos\alpha\cos\frac{\pi}{4}+\sin\alpha\sin\frac{\pi}{4}$
$[\cos(\text{A}-\text{B})=\cos\text{A}\cdot\cos\text{B}+\sin\text{A}\cdot\sin\text{B}]$
$\Rightarrow\cos\theta=\cos\alpha\cdot\frac{1}{\sqrt2}+\sin\alpha\cdot\frac{1}{\sqrt2}$
$\Rightarrow\sqrt2\cos\theta=\cos\alpha+\sin\alpha$
$\Rightarrow\sin\alpha+\cos\alpha=\sqrt2\cos\theta.$ Hence proved.

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