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Trigonometric Ratios — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Ratios5 Marks
Question
If $\sec\text{A}=\frac{5}{4},$ verify that $\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}.$
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Answer

Given: $\sec\text{A}=\frac{5}{4}\dots(1)$ To verify: $=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}\dots(2)$ Now we know that $\sec\text{A}=\frac{1}{\cos\text{A}}$ Therefore $\cos\text{A}=\frac{1}{\sec\text{A}}$ Now, by substituting the value of sec A from equation (1) We get, $\cos\text{A}=\frac{1}{\frac{5}{4}}$ $=\frac{4}{5}$ Therefore, $\cos\text{A}=\frac{4}{5}\dots(3)$ Now, we know the following trigonometric identity $\cos^2\text{A}+\sin^2\text{A}=1$ Therefore, $\sin^2\text{A}=1-\cos^2\text{A}$ Now by substituting the value of cos A from equation (3) We get, $\sin^2\text{A}=1-\Big(\frac{4}{5}\Big)^2$ $=1-\frac{\big(4\big)^2}{\big(5\big)^2}$ $=1-\frac{16}{25}$ Now by taking L.C.M We get, $\sin^2\text{A}=\frac{25-16}{25}$ $=\frac{9}{25}$ Now, by taking square root on both sides We get, $\sin\text{A}=\sqrt{\frac{9}{25}}$ $=\frac{\sqrt{9}}{\sqrt{25}}$ $=\frac{3}{5}$ Therefore, $\sin\text{A}=\frac{3}{5}\dots(4)$ Now, we know that $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}$ Now by substituting the value of sin A and cos A from equation (3) and (4) respectively We get, $\tan\text{A}=\frac{\frac{3}{5}}{\frac{4}{5}}$ $=\frac{3}{5}\times\frac{5}{4}$ $=\frac{3}{4}$ Therefore $\tan\text{A}=\frac{3}{4}\dots(5)$ Now from the expression of equation (2) $\text{L.H.S}=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}$Now by substituting the value of cos A and sin A from equation (3) and (4)
We get, $\text{L}.\text{H}.\text{S}=\frac{3\Big(\frac{3}{5}\Big)-4\Big(\frac{3}{5}\Big)^3}{4\Big(\frac{4}{5}\Big)^3-3\Big(\frac{4}{5}\Big)}$ Therefore, $\text{L}.\text{H}.\text{S}=\frac{\frac{9}{5}-4\Big(\frac{27}{125}\Big)}{4\Big(\frac{64}{125}\Big)-\frac{12}{5}}$ $=\frac{\frac{9}{5}-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}$ Now by taking L.C.M of both numerator and denominator We get, $\text{L}.\text{H}.\text{S}=\frac{\frac{9\times25}{5\times25}-\frac{108}{125}}{\frac{256}{125}-\frac{12\times25}{5\times25}}$ $=\frac{\frac{225}{125}-\frac{108}{125}}{\frac{256}{125}-\frac{300}{125}}$ $=\frac{\frac{225-108}{125}}{\frac{256-300}{125}}$ $=\frac{\frac{117}{125}}{\frac{-44}{125}}$ $=\frac{-117}{44}$ $=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{-117}{44}\dots(6)$ Now from the expression of equation (2) $\text{R}.\text{H}.\text{S}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}$ Now by substituting the value of tan A from equation (5) We get, $\text{R}.\text{H}.\text{S}=\frac{3\Big(\frac{3}{4}\Big)-\Big(\frac{3}{4}\Big)^3}{1-3\Big(\frac{3}{4}\Big)^2}$ $=\frac{\frac{9}{4}-\frac{27}{64}}{1-\frac{3\times9}{16}}$ Now by taking L.C.M We get, $\text{R}.\text{H}.\text{S}=\frac{\frac{9\times16}{4\times16}-\frac{27}{64}}{\frac{16-27}{16}}$ $=\frac{\frac{144}{64}-\frac{27}{64}}{\frac{-11}{16}}$ $=\frac{\frac{144-27}{64}}{\frac{-11}{16}}$ $=\frac{\frac{117}{64}}{\frac{-11}{16}}$ $=\frac{117}{64}\times\frac{16}{-11}$ $16\times4=64$ Now, Therefore, $\text{R}.\text{H}.\text{S}=\frac{117}{4}\times\frac{1}{-11}$ $=\frac{117\times1}{4\times-11}$ $=\frac{117}{-44}$ $=\frac{117}{-44}$ Therefore, $\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}=\frac{-117}{44}\dots(7)$ Now by comparing equation (6) and (7) $\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}$ We get, L.H.S. = R.H.S.

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