Ans: \[\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}\]

\[[0.\pi],x\neq\frac{\pi}{2}\]

Ans: \[\pi-\frac{\pi}{6}=\frac{5\pi}{6}\]

Ans: \[[\frac{\Pi}{2},\frac{3\Pi}{2}]\]

Ans: \[Tan^{-1}(-1)=-\frac{\pi}{4}\]

\[Cos^{-1}Cos(\pi+\frac{\pi}{6})=Cos^{-1}Cos(\pi-\frac{\pi}{6})\] \[=\frac{5\pi}{6}\]

Ans: \[\frac{2\pi}{3}+\frac{\pi}{3}=\pi\]

Ans: \[-\frac{\pi}{4}\]

Ans: 1 \[Because\:tan^{-1}x+tan^{-1}y=tan^{-1}{\frac{x+y}{1-xy}}\]

Ans: \[\frac{3\pi}{4}\] Principal interval of cot is \((0.\pi)\)  and \(-\frac{\pi}{4}\notin\...

Ans: \[\frac{\pi}{6}-\frac{\pi}{6}=0\]   We know that \(tan\frac{\pi}{6}=\frac{1}{\sqrt3}\) ...

Toolbox: The range of the principal value of $\tan^{-1}x$ is $\left [ -\large\frac{\pi}{2}...

Toolbox: The range of the principal value of $\; tan^{-1}x$ is $\left [ \large-\frac{\pi}{...

Toolbox: \( cos^{-1}x+cos^{-1}y=cos^{-1} (xy- \sqrt{1-x^2} \sqrt{1-y^2} )\) Given $cos^{-1} \la...

Toolbox: \(sin^{-1}(-\large\frac{1}{2})=-\large\frac{\pi}{6}\) \(sin\large\frac{\pi}{2}=1\) ...

A)    is the answer because a  .  b  =  ab cosø   which

A)  0      because  I.j  =  0  ,  j.k = 0 k.i...