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Chapter 19: Problem 1884

\(\sqrt{(3) \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=}\) (a) \(-4\) (b) 1 (c) 2 (d) 4

Short Answer

Expert verified
None of the given options (a, b, c, d) is the correct answer for the expression, as the final expression yields \( \sqrt{-\sin(5°)} \), which is not a real number.

Step by step solution

01

Rewrite the expression using trigonometric identities

Given expression: \( \sqrt{3 \operatorname{cosec}(20°) - \operatorname{sec}(20°)} \) Rewrite using identities: Expression becomes: \( \sqrt{3\frac{1}{\sin(20°)} - \frac{1}{\cos(20°)}} \)
02

Simplify the expression

We can simplify the expression by writing it in terms of common denominator: \( \sqrt{\frac{3\cos(20°) - \sin(20°)}{\sin(20°) \cos(20°)}} \) We know the identity: \( \sin(A) + \cos(B) = \sqrt{2}\left(\sin\!{\left(\!A + \frac{π}{4}\!\right)}\right) = \sqrt{2}\sin\!{\left(\!A + 45°\!\right)} \) We can rewrite the expression in the numerator as: \( 3\cos(20°) - \sin(20°) = \frac{1}{\sqrt{2}} [\sqrt{2} (3\cos(20°) - \sin(20°))] = \frac{1}{\sqrt{2}}[\sin(45° - 20°)] \) Now, we can rewrite the expression as: \( \sqrt{\frac{\frac{1}{\sqrt{2}}[\sin(45° - 20°)]}{\sin(20°) \cos(20°)}} \)
03

Cancel out the trigonometric terms

By using double angle formulas, we have: \(\sin 45°=\sqrt{2\sin 20°\cos 20°}\) Now, using this in the given expression: \( \sqrt{\frac{\frac{1}{\sqrt{2}}[\sin(45° - 20°)]}{\sin(20°) \cos(20°)}} = \sqrt{\frac{\frac{1}{\sqrt{2}}[\sin(45° - 20°)]}{\frac{1}{\sqrt{2}}\sin(45°)}} \) Cancelling out the terms, we get: \( \sqrt{\sin(45° - 20°)} \)
04

Find the value of the final expression

As we know that: \( \sin 45° = \frac{1}{\sqrt{2}} \) Now, replace the value in the expression: \( \sqrt{\sin(45° - 20°)} = \sqrt{\sin(25°)} = \sqrt{\sin(-365°)} \) As we know that sin(.) is periodic function of period 360, so \( \sin(-365°) = \sin(-365 + 360) = \sin(-5°) = -\sin(5°) \) So, the final expression becomes: \( \sqrt{-\sin(5°)} \) Since the square root of a negative number is not a real number, none of the given options are the correct answer for the expression.

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Most popular questions from this chapter

If \(\sin A=3 \sin (A+2 B)\) angle \(B\) is acute and \(A\) is obtuse: then (a) \(\tan \mathrm{B}=(1 / \sqrt{2})\) (b) \(\tan B>(1 / \sqrt{2})\) (c) \(\tan \mathrm{B}<(1 / \sqrt{2})\) (d) \(0<\tan B<(1 / \sqrt{2})\)

If \(\cos x+\cos y=0\) and \(\sin x+\sin y=0\) then \(\cos (x-y)=\) (a) 1 (b) \((1 / 2)\) (c) \(-1\) (d) \(-(1 / 2)\)

If \(\cos x=1-2 \sin ^{2} 32^{\circ}, \alpha, \beta\) are the value of \(x\) between \(0^{\circ}\) and \(360^{\circ}\) with \(\alpha<\beta\) then \(\alpha=\) (a) \(180^{\circ}-\beta\) (b) \(200^{\circ}-\beta\) (c) \((\beta / 4)-10^{\circ}\) (d) \((\beta / 5)-4^{\circ}\)

\(\sin ^{-1}(\sin 10)=\) (a) 10 (b) \(3 \pi-10\) (c) \(10-3 \pi\) (d) \(2 \pi-10\)

If \(\Delta \mathrm{ABC}, \mathrm{a}=2, \mathrm{~b}=3\) and \(\sin \mathrm{A}=(1 / 3)\), then \(\mathrm{B}=\) (a) \((\pi / 4)\) (b) \((\pi / 6)\) (c) \((\pi / 2)\) (d) \((\pi / 3)\)
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