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Q11:
Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$
Solution :
Step 1: Application of the Inverse Tangent Double Angle Identity
Given the equation: $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$
We apply the identity $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$, valid for $|\theta| < 1$. [Per the Properties of Inverse Trigonometric Functions]
Substituting $\theta = \cos x$, we get:
$\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}(2\csc x)$
Step 2: Algebraic Simplification
By applying the tangent function to both sides, we eliminate the inverse operators:
$\frac{2\cos x}{1-\cos^2 x} = 2\csc x$
Using the trigonometric identity $1-\cos^2 x = \sin^2 x$, we rewrite the denominator:
$\frac{2\cos x}{\sin^2 x} = 2 \cdot \frac{1}{\sin x}$ [Since $\csc x = \frac{1}{\sin x}$]
Dividing both sides by 2 and assuming $\sin x \neq 0$:
$\frac{\cos x}{\sin^2 x} = \frac{1}{\sin x}$
Step 3: Solving for $x$
Multiply both sides by $\sin^2 x$ (where $\sin x \neq 0$):
$\cos x = \sin x$
Divide by $\cos x$ (where $\cos x \neq 0$):
$\tan x = 1$
The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$. For $x \in [0, 2\pi]$, we consider the primary branch:
$x = \frac{\pi}{4}$
Step 4: Verification of Domain Constraints
We must ensure the arguments of the inverse functions are valid:
- For $\tan^{-1}(\cos x)$: $\cos(\pi/4) = \frac{1}{\sqrt{2}} < 1$. [Condition satisfied]
- For $\tan^{-1}(2\csc x)$: $\csc(\pi/4) = \sqrt{2}$. The argument is $2\sqrt{2}$. Since $\tan^{-1}$ is defined for all real numbers, this is valid.
Geometric Representation (Unit Circle)
Final Solution: $x = \frac{\pi}{4}$
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.MISC
- Q1: Find the value of the following: $\cos^{-1}(\cos(\frac{13\pi}{6}))$
- Q2: Find the value of the following: $\tan^{-1}(\tan(\frac{7\pi}{6}))$
- Q3: Prove that $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
- Q4: Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
- Q5: Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
- Q6: Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
- Q7: Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
- Q8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x})$, $x \in [0, 1]$
- Q9: Prove that $\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}) = \frac{x}{2}$, $x \in (0, \frac{\pi}{4})$
- Q10: Prove that $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- Q12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x > 0)$
- Q13: $\sin(\tan^{-1} x)$, $|x| < 1$ is equal to
- Q14: $\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
Chapters in CBSE - Class 12 Mathematics
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