Inverse Trigonometric Functions

### Mark the tick against the correct answer in the following: Domain of is A. B. C. D. Solution: Option(C) is correct. To Find: The Domain of $\sec ^{-1}(x)$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\sec ^{-1}(x)$ can be obtained from the graph...

### Mark the tick against the correct answer in the following: Domain of is A. B. C. D. None of these

Solution: Option(B) is correct. To Find: The Domain of $\cos ^{-1}(x)$ Here,the inverse function of $\cos$ is given by $y=f^{-1}(x)$ The graph of the function $y=\cos ^{-1}(x)$ can be obtained from...

### Mark the tick against the correct answer in the following: Range of is A. B. C. D. None of these

Solution: Option(C) is correct. To Find: The range of $\operatorname{cosec}^{-1}(x)$ Here,the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function...

### Mark the tick against the correct answer in the following: Range of is A. B. C. D. None of these

Solution: Option(C) is correct. To Find:The range of $\sec ^{-1}(x)$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\sec ^{-1}(x)$ can be obtained from the graph of...

### Mark the tick against the correct answer in the following: Range of is A. B. C. D. None of these

Solution: Option(B) is correct. To Find: The range of $\tan ^{-1} x$ Here, the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\tan ^{-1}(x)$ can be obtained from the graph...

### Mark the tick against the correct answer in the following: Range of is A. B. C. D. None of these

Solution: Option(A) is correct. To Find: The range of $\cos ^{-1} x$ Here, the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function $y=\cos ^{-1}(x)$ can be...

### Mark the tick against the correct answer in the following: Range of is A. B. C. D. None of these

Solution: Option() is correct. To Find: The range of $\sin ^{-1} x$ Here,the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function $y=\sin ^{-1}(x)$ can be obtained...

### Mark the tick against the correct answer in the following: A. B. C. D. Solution: Option(B) is correct. To Find: The value of $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ Now $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ can be written in...

### Mark the tick against the correct answer in the following: A. B. C. D. Solution: Option(A) is correct. To Find: The value of $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$ Let, $x=\cos ^{-1} \frac{\sqrt{5}}{3}$ $\Rightarrow \cos x=\frac{\sqrt{5}}{3}$ Now,...

### Mark the tick against the correct answer in the following: The principal value of is A. B. C. D. Solution: Option(B) is correct. To Find: The Principle value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by $x$ Now, let $x=\cos ^{-1}\left(\frac{-1}{2}\right)$...

### Mark the tick against the correct answer in the following: The principal value of is A. B. C. D. none of these

Solution: Option(A) is correct. To Find: The Principle value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by $x$ Now, let $x=\sin ^{-1}\left(\frac{-1}{2}\right)$...

### Prove that: Solution: To Prove: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$ Proof:...

### Prove that: Solution: To Prove: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{-1} \frac{4}{3} \Rightarrow 2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3}$...

### Prove that: Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof:...

### Prove that: Solution: To Prove: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof:...

### Prove that: Solution: To Prove: $\tan ^{-1}\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{x y}}\right)=\tan ^{-1} \sqrt{x}+\tan ^{-1} \sqrt{y}$ We know that, $\tan A+\tan B=\frac{\tan A+\tan B}{1-\tan A \tan B}$ Also,...

### Prove that: Solution: To Prove: $\cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$ Formula Used: 1) $\tan \left(\frac{\pi}{4}+A\right)=\frac{1+\tan A}{1-\tan A}$ 2)...

### Prove that: Solution: To Prove: $\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=2 \cos ^{-1} x$ Formula Used: 1) $\cos 2 A=2 \cos ^{2} A-1$ 2) $\cos ^{-1} A=\sec ^{-1}\left(\frac{1}{A}\right)$ Proof:...

### Prove that: Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...

### Find the principal value of each of the following : Solution: $\tan ^{-1}(-\sqrt{3})=-\tan ^{-1}(\sqrt{3})$ [Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$ $=-\frac{\pi}{3}$

### Find the principal value of each of the following : Solution: $\cot ^{-1}(-1)=\pi-\cot ^{-1}(1)$ [Formula: $\left.\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)\right]$ $\begin{array}{l} =\pi-\frac{\pi}{4} \\ =\frac{3 \pi}{4} \end{array}$

### Find the principal value of each of the following : Solution: $\operatorname{cosec}^{-1}(-\sqrt{2})=-\operatorname{cosec}^{-1}(\sqrt{2})$ [Formula: $\left.\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)\right]$ $=-\frac{\pi}{4}$ This can...

### Find the principal value of each of the following : Solution: $\sec ^{-1}(-2)=\pi-\sec ^{-1}(2)\left[\right.$ Formula: $\left.\sec ^{-1}(-x)=\pi-\sec ^{-1}(x)\right]$ $\begin{array}{l} =\pi-\frac{\pi}{3} \\ =\frac{2 \pi}{3} \end{array}$

### Find the principal value of each of the following : Solution: $\tan (-1)=-\tan (1)\left[\right.$ Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$ [ We know that $\tan \frac{\pi}{4}=1$, thus $\left.\tan ^{-1} \frac{\pi}{4}=1\right]$...

### Find the principal value of each of the following : Solution: $\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right)$ [ Formula: $\left.\cos ^{-1}(-x)=-\cos ^{-1}(x)\right]$ $=\pi-\frac{\pi}{3}$ $=\frac{2 \pi}{3}$

### Find the principal value of each of the following : Solution: $\sin ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)\left[\right.$ Formula: $\left.\sin ^{-1}(-x)=\sin ^{-1}(x)\right]$ $=-\frac{\pi}{6}$

### Evaluate $\begin{array}{l} \sin \left(\frac{\pi}{2}+\frac{\pi}{3}\right) \\ =\sin \left(\frac{5 \pi}{6}\right) \\ =\sin \left(\pi-\frac{\pi}{6}\right) \\ =\sin \frac{\pi}{6} \\ =\frac{1}{2} \end{array}$

### Evaluate $\cos \left\{\pi-\frac{\pi}{6}+\frac{\pi}{6}\right\}$ $\begin{array}{l} =\cos \{\pi\} \\ =\cos \left(\frac{\pi}{2}+\frac{\pi}{2}\right) \\ =-1 \end{array}$