The solution of the equation sec theta - {rm{ | vedclass

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Question

(a) $3(\sin 	heta - \cos 	heta ) = 4\sin 	heta \cos 	heta $

==> $3(\sin 	heta - \cos 	heta ) = 2\sin 2	heta $

Squaring both sides, we

Vedclass · Accepted Answer

$\frac{1}{2}[n\pi + {( - 1)^n}{\sin ^{ - 1}}(3/4)]$

Vedclass · Answer

(a) $3(\sin 	heta - \cos 	heta ) = 4\sin 	heta \cos 	heta $

==> $3(\sin 	heta - \cos 	heta ) = 2\sin 2	heta $

Squaring both sides, we get $9(1 - S) = 4{S^2},$

where $S = \sin 2	heta $ or $4{S^2} + 9S - 9 = 0$.

$	herefore $ $\,(S + 3)\,(4S - 3) = 0$ or $S = \frac{3}{4}$ as $S 
e - 3$

or $\sin 2	heta = \frac{3}{4} = \sin \alpha $

$	herefore $ $2	heta = n\pi + {( - 1)^n}\alpha $

or $	heta = \frac{1}{2}\,\left[ {n\pi + {{( - 1)}^n}{{\sin }^{ - 1}}\left( {\frac{3}{4}} ight)} ight]$.

The solution of the equation $\sec \theta - {\rm{cosec}}\theta = \frac{4}{3}$ is

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