Question

$if \: \alpha \: belongs \: to \: ({\pi \: \frac{3\pi}{2} }) \: then \: { \tan }^{ - 1}( \tan( \alpha ))$
is equal to



​

Answer 1

Answer:

if \: \alpha \: belongs \: to \: ({\pi \: \frac{3\pi}{2} }) \: then \: { \tan }^{ - 1}( \tan( \alpha ))ifαbelongsto(π

2

3π

)thentan

−1

(tan(α))

is equal to

Answer 2

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

Given inverse Trigonometric function is

$\rm :\longmapsto\: {tan}^{ - 1} (tan \alpha )$

and

$\rm :\longmapsto\: \alpha \: \in \: \bigg(\pi, \: \dfrac{3\pi}{2} \bigg)$

We know,

$\boxed{\tt{ {tan}^{ - 1} (tanx) = x \: \: \: if \: x \: \in \: \bigg( - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg) }}$

So,

$\rm :\longmapsto\: \alpha \: \cancel \in \: \bigg( - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg)$

Now,

$\rm :\longmapsto\:\pi < \alpha < \dfrac{3\pi}{2}$

$\rm :\longmapsto\: - \pi > - \alpha > - \dfrac{3\pi}{2}$

$\rm :\longmapsto\:\pi - \pi >\pi - \alpha > \pi - \dfrac{3\pi}{2}$

$\rm :\longmapsto\:0 > \pi - \alpha > - \dfrac{\pi}{2}$

$\rm :\longmapsto\: - \dfrac{\pi}{2} < \pi - \alpha < 0$

Now,

$\rm :\longmapsto\:tan(\pi - \alpha ) = - tan \alpha$

$\bf\implies \:tan \alpha = - tan(\pi - \alpha ) = tan( \alpha - \pi)$

Therefore,

$\rm :\longmapsto\: {tan}^{ - 1} (tan \alpha )$

$\rm \:  =  \: {tan}^{ - 1} [tan( \alpha - \pi)]$

$\rm \:  =  \: \alpha - \pi$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \:  {tan}^{ - 1} (tan \alpha ) =  \: \alpha - \pi \: }}$

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MORE TO KNOW

$\boxed{\tt{ {tan}^{ - 1} (tanx) = x \: \: \: if \: x \: \in \: \bigg( - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg) }}$

$\boxed{\tt{ {sin}^{ - 1} (sinx) = x \: \: \: if \: x \: \in \: \bigg[- \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg] }}$

$\boxed{\tt{ {cos}^{ - 1} (cosx) = x \: \: \: if \: x \: \in \: \bigg[0, \: \pi\bigg] }}$