Question

The principal value of ${{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)$ is
$A.\dfrac{-3\pi }{4}$
$B.\dfrac{3\pi }{4}$
$C.\dfrac{-\pi }{4}$
$D.\dfrac{\pi }{4}$

Answer 1

Hint: In the question, we need to find the principal value of the function ${{\tan }^{-1}}$. The principal value of ${{\tan }^{-1}}(x)$ is $\theta$ if $\dfrac{-\pi }{2}<\theta <\dfrac{\pi }{2}$ and its general value is $n\pi +\theta$. To solve use basic trigonometry identities formula like $\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta$

Complete step-by-step answer:
From the question, we can rewrite it as

${{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)={{\tan }^{-1}}\left( cot\left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right) \right)$
Now, we will apply the identity $\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta$ in the above expression. Then, we will get

& \Rightarrow {{\tan }^{-1}}\left( -\tan \dfrac{\pi }{4} \right) \\\ & \because \tan \left( -\theta \right)=-\tan \theta \\\ & \Rightarrow {{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{4} \right) \right) \\\ \end{aligned}$$ Now, we know that $${{\tan }^{-1}}\left( \tan \theta \right)=\theta $$, so we will use it and get $$\Rightarrow \left( -\dfrac{\pi }{4} \right)$$ Hence, the principal value of $${{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)$$ is $$\dfrac{-\pi }{4}$$ . Principal value for tan function always lies between $$\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$$. So, if somehow the obtained value is less than $$-\dfrac{\pi }{2}$$ or greater than $$\dfrac{\pi }{2}$$, then we must have to bring the principal value in between $$\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$$ using trigonometric identities as explained above. Therefore, option C is the correct one. Note: Sometimes the question would be like finding the principal value of $$\left( {{\tan }^{-1}}\left( -\dfrac{\pi }{6} \right) \right)$$. For this question, be careful, because in this question $$\dfrac{-\pi }{6}$$ is lying between $$\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$$. Therefore, its principal value would be directly $$\dfrac{-\pi }{6}$$. If $$\dfrac{-\pi }{6}$$wouldn’t be lying between $$\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$$. Then, $$\dfrac{-\pi }{6}$$ wouldn’t be the principal value of this question. For example: for the question $${{\tan }^{-1}}\left( \dfrac{5\pi }{3} \right)$$ , principal value wouldn’t be $$\dfrac{5\pi }{3}$$. To find its principal value, we need to use the general formula $$n\pi +\theta $$. In the above question $${{\tan }^{-1}}\left( cot\dfrac{3\pi }{4} \right)$$, to find its principal value, in between we used identity $$\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta $$. But this formula isn't the same for $$\cot \left( \dfrac{\pi }{2}-\dfrac{\pi }{3} \right)$$. For this expression, the identity is $$\cot \left( \dfrac{\pi }{2}-x \right)$$$$=\tan x$$. So, be careful while using identity.So, for$${{\tan }^{-1}}\left( \cot \left( \dfrac{\pi }{2}-\dfrac{\pi }{3} \right) \right)$$ ,the principal value would be $$\dfrac{\pi }{3}$$ .