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Trigonometric Ratios — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Ratios3 Marks
Question
Evaluate the following:
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ$
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Answer

We have,
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ\dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\cos60^\circ=\frac{1}{2},\sec45^\circ=\sqrt{2},\sec30^\circ=\frac{2}{\sqrt{3}}$
So by substituting above values in equation (1)
We get,
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ$
$=\big(\sqrt{3}\big)^2-2\Big(\frac{1}{2}\Big)^2-\frac{3}{4}\big(\sqrt{2}\big)^2-4\Big(\frac{2}{\sqrt{3}}\Big)^2$
$=3-2\times\frac{1^1}{2^2}-\frac{3}{4}\times2-4\times\frac{2^2}{\big(\sqrt{3}\big )^2}$
Now, in the third term 4 gets cancelled by 2 and 2 remains
Therefore,
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ$
$=3-2\times\frac{1}{4}-\frac{3}{2}-4\times\frac{4}{3}$
Now in the second term, 4 gets cancelled by 2 and 2 remains
Therefore,
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ$
$=3-2\times\frac{1}{4}-\frac{3}{2}-4\times\frac{4}{3}$
$=3-\frac{1}{2}-\frac{3}{2}-4\times\frac{4}{3}$
$=3-\frac{1}{2}-\frac{3}{2}-\frac{16}{3}$
Now, LCM of denominator in the above expression is 6
Therefore by taking LCM
We get,
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ$
$=\frac{3\times6}{1\times6}-\frac{1\times3}{2\times3}-\frac{3\times3}{2\times3}-\frac{16\times2}{3\times2}$
$=\frac{18}{6}-\frac{3}{6}-\frac{9}{6}-\frac{32}{6}$
$=\frac{18-3-9-32}{6}$
$=\frac{18-12-32}{6}$
$=\frac{18-44}{6}$
$=\frac{-26}{6}$
Now in the above expression, $=\frac{-26}{6}$ gets reduced to $\frac{-13}{3}$
$\cot^230^\circ-2\cos^260^\circ-\frac{3}{4}\sec^245^\circ-4\sec^230^\circ=\frac{-13}{3}$

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