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Trigonometric Ratios — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Ratios3 Marks
Question
If $\sec\text{A}=\frac{17}{8},$ verify that $\frac{3-4\sin^2\text{A}}{4\cos^2\text{A}-3}=\frac{3-\tan^2\text{A}}{1-3\tan^2\text{A}}.$
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Answer

Given:
$\sec\text{A}=\frac {17}{8}\dots(1)$
To verify:
$\frac{3- 4\sin^2\text{A}}{4\cos^2\text{A}-3}=\frac{3-\tan^2\text{A}}{1-3\tan^2\text{A}}\dots(2)$
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,
$\cos\text{A}=\frac {1}{\frac{17}{8}}$
$=\frac{8}{17}$
Therefore,
$\cos\text{A}=\frac {8}{17}\dots(3)$
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,
$\sin^2\text{A}=1- \Big(\frac{8}{17}\Big)^2$
$=1-\frac{(8)^2} {(17)^2}$
$=1-\frac{64} {289}$
Now by taking L.C.M.
We get,
$\sin^2\text{A}= \frac{289-64}{289}$
$=\frac{225}{289}$
Now, by taking square root on both sides
We get,
$\sin\text{A}=\sqrt {\frac{225}{289}}$
$=\frac{\sqrt{225}} {\sqrt{289}}$
$=\frac{15}{17}$
Therefore,
$\sin\text{A}=\frac{15} {17}\dots(4)$
Now, we know that $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}$
Now by substituting the value of cos A and sin A from equation (3) and (4) respectively
We get,
$\tan\text{A}=\frac {\frac{15}{17}}{\frac{8}{17}}$
$=\frac{15} {17}\times\frac{17}{8}$
$=\frac{15}{8}$
Therefore
$\tan\text{A}=\frac {15}{8}\dots(5)$
Now from the expression of equation (2)
$\text{L}.\text{H}. \text{S}=\frac{3-4\sin^2\text{A}}{4\cos^2\text{A}-3}$
Now by substituting the value of cos A and sin A from equation (3) and (4)
We get,
$\text{L}.\text{H}.\text {S}=\frac{3-4\Big(\frac{15}{17}\Big)^2}{4\Big(\frac{8}{17}\Big)^2-3}$
Therefore,
$\text{L}.\text{H}. \text{S}=\frac{3-4\Big(\frac{225}{289}\Big)}{4\Big(\frac{64}{289}\Big)-3}$
$=\frac{3-\frac {900}{289}}{\frac{256}{289}-3}$
Now by taking L.C.M of both numerator and denominator
We get,
$\text{L}.\text{H}. \text{S}=\frac{\frac{3\times289}{1\times289}-\frac{900}{289}}{\frac{256}{289}-\frac{3\times289} {1\times289}}$
$=\frac{\frac{867} {289}-\frac{900}{289}}{\frac{256}{289}-\frac{867}{289}}$
$=\frac{\frac{867- 900}{289}}{\frac{256-867}{289}}$
$=\frac{\frac{-33} {289}}{\frac{-611}{289}}$
$=\frac{33}{611}$
Therefore,
$\frac{3- 4\sin^2\text{A}}{4\cos^2\text{A}-3}=\frac{33}{611}\dots(6)$
Now from the expression of equation (2)
$\text{R}.\text{H}. \text{S}=\frac{3-\tan^2\text{A}}{1-3\tan^2\text{A}}$
Now by substituting the value of tan A from equation (5)
We get,
$\text{R}.\text{H}. \text{S}=\frac{3-\Big(\frac{15}{8}\Big)^2}{1-3\Big(\frac{15}{8}\Big)^2}$
$=\frac{3-\frac{225}{64}}{1-\frac{3\times225}{64}}$
Now by taking L.C.M
We get,
$\text{R}.\text{H}. \text{S}=\frac{\frac{3\times64}{1\times64}-\frac{225}{64}}{\frac{64-675}{64}}$
$=\frac{\frac{192} {64}-\frac{225}{64}}{\frac{-611}{64}}$
$=\frac{\frac{192- 225}{64}}{\frac{-611}{64}}$
Therefore
$\text{R}.\text{H}. \text{S}=\frac{\frac{-33}{64}}{\frac{-611}{64}}$
$=\frac{-33} {64}\times\frac{64}{-611}$
$=\frac{33}{611}$
Therefore,
$\frac{3-\tan^2\text {A}}{1-3\tan^2\text{A}}=\frac{33}{611}\dots(7)$
Now by comparing equation (6) and (7)
We get,
$\frac{3- 4\sin^2\text{A}}{4\cos^2\text{A}-3}=\frac{3-\tan^2\text{A}}{1-3\tan^2\text{A}}$

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