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Trigonometric Ratios Of Compound Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles2 Marks
Question
Prove that:
$\frac{\cos2\text{x}}{1+\sin2\text{x}}=\tan(\frac{\pi}{4}-\text{x})$
✓

Answer

$\text{LHS}=\frac{\cos2\text{x}}{1+\sin2\text{x}}$ $=\frac{\cos^2\text{x}-\sin^2\text{x}}{\sin^2+\cos^2\text{x}+2\sin\text{x}\cos\text{x}}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}\ \&\sin^2\text{x}+\cos^2\text{x}=1]$ $=\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}+\sin\text{x})}{(\cos\text{x}+\sin\text{x})^2}$ $[\because\text{a}^2-\text{b}^2=(\text{a+b})(\text{a-b})]$ $=\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}$ Dividing num erator and demom enator by $\cos\text{x}$ $=\frac{1-\tan\text{x}}{1+\tan\text{x}}$ $=\tan(\frac{\pi}{4}-\text{x})=\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$ $=\frac{1-\tan\text{x}}{1+\tan\text{x}}\ \text{RHS}$Note: $\tan(\frac{\pi}{4}-\text{x})=\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$
$=\frac{1-\tan\text{x}}{1+\tan\text{x}}$

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