Question
Solve the following equation for x:
$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sin\text{x}),\text{x}\neq\frac{\pi}{2}.$
$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sin\text{x}),\text{x}\neq\frac{\pi}{2}.$
✓
Answer
$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sec\text{x})$
$\tan^{-1}\Big(\frac{2\sin\text{x}}{1-\sin^{2}\text{x}}\Big)=\tan^{-1}(2\sec\text{x})$ $\Big[\text{Since }2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\frac{2\sin\text{x}}{\cos^2\text{x}}=2\sec\text{x}$
$\frac{\sin\text{x}}{\cos\text{x}.\cos\text{x}}=\sec\text{x}$
$\tan\text{x}\sec\text{x}=\sec\text{x}$
$\tan\text{x}=1$
$\text{x}=\frac{\pi}{4}$
Hence the value of x is $\frac{\pi}{4}$
Thus, the solution is $\text{x}=\text{n}\pi+\frac{\pi}{4}$
$\tan^{-1}\Big(\frac{2\sin\text{x}}{1-\sin^{2}\text{x}}\Big)=\tan^{-1}(2\sec\text{x})$ $\Big[\text{Since }2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\frac{2\sin\text{x}}{\cos^2\text{x}}=2\sec\text{x}$
$\frac{\sin\text{x}}{\cos\text{x}.\cos\text{x}}=\sec\text{x}$
$\tan\text{x}\sec\text{x}=\sec\text{x}$
$\tan\text{x}=1$
$\text{x}=\frac{\pi}{4}$
Hence the value of x is $\frac{\pi}{4}$
Thus, the solution is $\text{x}=\text{n}\pi+\frac{\pi}{4}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Let $f : R → R$ be defined as $\text{f(x)}=\frac{2\text{x}-3}{4}.$ Write $fof ^{-1}(1)$.
→Solve the following by inversion method $2x + y = 5, 3x + 5y = −3$
→Express the following switching circuit in symbolic form of logic. Construct its switching table and write your conclusion from it :
Evaluate the following integrals:
$\int\frac{\log\text{x}^2}{\text{x}}\text{dx}$
→$\int\frac{\log\text{x}^2}{\text{x}}\text{dx}$
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
→$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Find the distance of the point $4 \hat{i}-3 \hat{j}+\hat{k}$ from the plane $\bar{r} \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})=21$.
→Find A (adjoint A) for the matrix $\text{A}=\begin{bmatrix}1 & -2 & 3 \\0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$.
→Let $\text{A}=\begin{bmatrix}-1&0&2\\3&1&4 \end{bmatrix},\text{ B}=\begin{bmatrix}0&-2&5\\1&-3&1 \end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&-5&2\\6&0&-4 \end{bmatrix}.$ Compute 2A - 3B + 4C.
→Find the probability distribution of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.
Show that the lines $\frac{\text{x}-5}{7}=\frac{\text{y}+2}{-5}=\frac{\text{z}}{1}$ and $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ are perpendicular to each other.
→