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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:
$\frac{\cos\text{x}}{1-\sin\text{x}}=\tan(\frac{\pi}{4}+\frac{\pi}{2})$
✓

Answer

$\text{LHS}=\frac{\cos\text{x}}{1-\sin\text{x}}$
$\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}-2\sin\frac{\text{x}}{2},\frac{\cos\text{x}}{2}}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin2\text{x}\ \&\sin^2\text{x}+\cos^2\text{x}]=1$
$\frac{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2})}{(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2})}$
$=\frac{\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}}$
Dividing numeretor and denominator by $\cos\frac {\text{x}}{2}$
$=\frac{1+\tan\frac{\text{x}}{2}}{1=\tan\frac{\text{x}}{2}}$
$\tan(\frac{\pi}{4}+\frac{\text{x}}{2})=\text{RHS}$

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