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3.Trigonometrical Ratios, Functions and Identities
hard
જો $0 < x < \pi $ અને $\cos x + \sin x = \frac{1}{2}$ તો $\tan x$ મેળાવે. . .
A
$\frac{{1 - \sqrt 7 }}{4}$
B
$\;\frac{{4 - \sqrt 7 }}{3}$
C
$ - \frac{{4 + \sqrt 7 }}{3}$
D
$\;\frac{{1 + \sqrt 7 }}{4}$
(AIEEE-2006)
Solution
$\cos x+\sin x=\frac{1}{2}$
$\Rightarrow(\cos x+\sin x)^{2}=\frac{1}{4}$
$\Rightarrow \cos ^{2} x+\sin ^{2} x+2 \cos x \sin x=\frac{1}{4}$
$\left[\because \cos ^{2} x+\sin ^{2} x=1 \text { and } 2 \cos x \sin x=\sin 2 x\right]$
$\Rightarrow 1+\sin 2 x=\frac{1}{4}$
$\Rightarrow \sin 2 x=-\frac{3}{4},$ so $x$ is obtuse and
$\frac{2 \tan x}{1+\tan ^{2} x}=-\frac{3}{4}$
$\Rightarrow 3 \tan ^{2} x+8 \tan x+3=0$
$\tan x=\frac{-8+\sqrt{64-36}}{6}$
$=\frac{-4+\sqrt{7}}{3}$
as $\tan x<0$
$\tan x=\frac{-4-\sqrt{7}}{3}$
Std 11
Mathematics