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Trigonometrical Equations
hard
સમીકરણ $\sec \theta - {\rm{cosec}}\theta = \frac{4}{3}$ ઉકેલ મેળવો.
A
$\frac{1}{2}[n\pi + {( - 1)^n}{\sin ^{ - 1}}(3/4)]$
B
$n\pi + {( - 1)^n}{\sin ^{ - 1}}(3/4)$
C
$\frac{{n\pi }}{2} + {( - 1)^n}{\sin ^{ - 1}}(3/4)$
D
એકપણ નહિ.
Solution
(a) $3(\sin \theta – \cos \theta ) = 4\sin \theta \cos \theta $
==> $3(\sin \theta – \cos \theta ) = 2\sin 2\theta $
Squaring both sides, we get $9(1 – S) = 4{S^2},$
where $S = \sin 2\theta $ or $4{S^2} + 9S – 9 = 0$.
$\therefore $ $\,(S + 3)\,(4S – 3) = 0$ or $S = \frac{3}{4}$ as $S \ne – 3$
or $\sin 2\theta = \frac{3}{4} = \sin \alpha $
$\therefore $ $2\theta = n\pi + {( – 1)^n}\alpha $
or $\theta = \frac{1}{2}\,\left[ {n\pi + {{( – 1)}^n}{{\sin }^{ – 1}}\left( {\frac{3}{4}} \right)} \right]$.
Standard 11
Mathematics
