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Trigonometric Identities — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
Question
If $\text{m}=(\cos\theta-\sin\theta)$ and $\text{n}=(\cos\theta+\sin\theta),$ then show that $\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}=\frac{2}{\sqrt{1-\tan^2\theta}}.$
✓

Answer

$\text{LHS}=\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}$$=\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}$
$=\frac{\text{m}+\text{n}}{\sqrt{\text{mn}}}$
$=\frac{(\cos\theta-\sin\theta)+(\cos\theta+\sin\theta)}{\sqrt{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}}$
$=\frac{2\cos\theta}{\sqrt{\cos^2\theta-\sin^2\theta}}$
$=\frac{\big(\frac{2\cos\theta}{\cos\theta}\big)}{\bigg(\frac{\sqrt{cos^2\theta-\sin^2\theta}}{\cos\theta}\bigg)}$
$=\frac{2}{\sqrt{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}}$
$=\frac{2}{1-\tan^2\theta}$
$=\text{RHS}$

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