Question

find the general solution of
tan theta = -1​

Answer 1

$\large{\underline{\underline{\sf{\purple{Given}}}}}$

• tan theta = -1

$\large{\underline{\underline{\sf{\purple{To\: find }}}}}$

• General solution of given equation.

$\large{\underline{\underline{\sf{\purple{Solution}}}}}$

$\sf{\implies \tan \theta = -1}$

$\sf{\implies - 1 = - \tan (\frac{\pi}{4}) }$

$\sf{\implies - \tan( \frac{\pi}{4}) = \tan ( - \frac{\pi}{4} )}$

$\sf{\implies \tan ( \frac{-\pi }{4}) = \tan ( \pi - \frac{ \pi}{4} ) }$

$\sf{\implies \tan ( \pi - \frac{ \pi}{4} ) = \tan ( \frac{3 \pi}{4}) }$

$\sf{\implies \tan \theta = \tan ( \frac{ 3 \pi}{4}) }$

$\sf{\implies \theta = \frac{ 3 \pi}{4} }$

We already know it will be written as :-

$\sf{\implies \theta = n \pi + \alpha , n \: belongs \: to \: Z}$

${\underline{\sf{\implies \theta = n \pi + \frac{3 \pi}{4} , n \; belongs\: to \: Z}}}$

Other important identities :-

Trigonometric equation = General equation .

$\sf{\implies \sin \theta = \sin \alpha ;\: GS \: \theta = n \pi + \alpha + {-1}^{n} \alpha , n \: belongs\: to \: Z}$

$\sf{\implies \cos \theta = \cos \alpha ; \: GS \: \theta = 2n \pi \pm \alpha , n \: belongs\: to \: Z}$

$\sf{\implies \sin \theta = 0 \: , GS\: \: theta = n \pi ,\: n \:belong \:to\: Z }$

$\sf{\implies \cos \theta = 0 , \: GS \: \theta = (2n +1 ) \frac{\pi}{2} , n \: belongs\: to \: Z }$

$\sf{\implies \tan \theta = 0 , \: GS\: \theta = n \pi , \: n\: belongs\: to \: Z }$