Question

solve it please.....​

Answer 1

Here you go

$tan^-^1(tan \frac{3\pi}{4} )$

$\frac{3\pi}{4}$ is in 2nd quadrant

So, the function will be $\pi-\theta$

and the value of tanθ will be negative

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

the range of tanθ is $(\frac{-\pi}{2} ,\:\frac{\pi}{2} )$

so, the answer is $\frac{-\pi}{4}$

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Answer 2

Answer:

tan−1(tan(3π/4))

=tan−1(tan(π−π/4))

=tan−1(tan(−π/4)

=−π/4

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