JEE Main 2021Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestionIf π2≤x≤3π4\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}2π≤x≤43π, then cos−1(1213cosx+513sinx)\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)cos−1(1312cosx+135sinx) is equal toOptionsAx+tan−1512x+\tan ^{-1} \frac{5}{12}x+tan−1125Bx−tan−143x-\tan ^{-1} \frac{4}{3}x−tan−134Cx+tan−145x+\tan ^{-1} \frac{4}{5}x+tan−154Dx−tan−1512x-\tan ^{-1} \frac{5}{12}x−tan−1125Check AnswerHide SolutionSolutionπ2≤x≤3π4cos−1(1213cosx+512sinx)cos−1(cosxcosα+sinxsinα)cos−1(cos(x−α))⇒x−α because x−α∈(−π2,π2)⇒x−tan−1512\begin{aligned} & \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4} \\ & \cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{12} \sin x\right) \\ & \cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha) \\ & \cos ^{-1}(\cos (x-\alpha)) \\ & \Rightarrow x-\alpha \text { because } x-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \Rightarrow x-\tan ^{-1} \frac{5}{12} \end{aligned}2π≤x≤43πcos−1(1312cosx+125sinx)cos−1(cosxcosα+sinxsinα)cos−1(cos(x−α))⇒x−α because x−α∈(−2π,2π)⇒x−tan−1125