Top 50 Most Repeated INVERSE TRIGONOMETRIC FUNCTIONS PYQs | JEE MAINS
A curated collection of the most important questions from INVERSE TRIGONOMETRIC FUNCTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from INVERSE TRIGONOMETRIC FUNCTIONS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →Domain of sin⁻¹ is [-1,1], so -1 ≤ 2x−1 ≤ 1 → 0 ≤ x ≤ 1 → length = 1....
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy>1$, result is $\frac{3\pi}{4}$....
Read Full Step-by-Step Solution →The function $ y = \cos^{-1}(x) $ has domain [-1,1] and range [0,\pi]. It is neither even nor odd. However, it is symmetric about the point (0, \pi/2)...
Read Full Step-by-Step Solution →Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, solve $\frac{x+y}{1-1/7} = 1$ → $x+y=1$....
Read Full Step-by-Step Solution →$\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\tan^{-1}(1) = \pi/4$, sum = $\pi/2 \approx 1.57$....
Read Full Step-by-Step Solution →$\sin^{-1}(\sin x)$ is periodic with linear segments due to principal value restriction, forming a sawtooth pattern....
Read Full Step-by-Step Solution →Identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \approx 1.57$, independent of $x$ in domain....
Read Full Step-by-Step Solution →Using the identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, we have $\cos^{-1}(3/5)=\frac{\pi}{2}-\theta$. $\theta=\sin^{-1}(0.6)\approx0.644\,\text{rad...
Read Full Step-by-Step Solution →