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

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles3 Marks
Question
If $\tan\text{A}=\frac{1}{7}$ and $\tan\text{B}=\frac{1}{3},$ show that $\cos2\text{A}=\sin4\text{B}$
✓

Answer

We have,
$\tan\text{A}=\frac{1}{7}\&\tan\text{B}=\frac{1}{3}$
$\therefore\cos2\text{A}=\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\frac{1-\Big(\frac{1}{7}\Big)^2}{1+\Big(\frac{1}{7}\Big)^2}=\frac{48}{\frac{49}{\frac{50}{49}}}$
$=\frac{48}{50}=\frac{24}{25}\ ...{\text{(A)}}$
Also,
$\sin4\text{B}=\sin2.2.\text{B}$
$=2\sin2\text{B}.\cos2\text{B}$
$=2.\Big(\frac{2\tan\text{B}}{1+\tan^2\text{B}}\Big).\Big(\frac{1-\tan^2\text{B}}{1+\tan^2\text{B}}\Big)$
$=4.\Bigg(\frac{\frac{1}{3}}{1+\frac{1}{9}}\Bigg).\Bigg(\frac{1-\frac{1}{9}}{1+\frac{1}{9}}\Bigg)$
$=\frac{4.\frac{1}{3}.\frac{8}{9}}{\frac{10}{9}\times\frac{10}{9}}$
$=\frac{32\times3}{100}$
$=\frac{8\times3}{25}=\frac{24}{25}\ ...(\text{B})$
From(A) & (B)
$\cos2\text{A}=\sin4\text{B}$

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