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Q6:
Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
Solution :
Step 1: Conversion of Inverse Trigonometric Functions to a Common Domain
To evaluate the sum $\cos^{-1} \left( \frac{12}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right)$, it is most efficient to express both terms as $\sin^{-1}$ functions. [Per the identity: $\cos^{-1}(x) = \sin^{-1}(\sqrt{1-x^2})$ for $x \in [0, 1]$].
Let $\alpha = \cos^{-1} \left( \frac{12}{13} \right)$. Then, $\cos \alpha = \frac{12}{13}$.
Using the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$:
$\sin \alpha = \sqrt{1 - \left( \frac{12}{13} \right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.
Therefore, $\alpha = \sin^{-1} \left( \frac{5}{13} \right)$.
Step 2: Substitution and Algebraic Application
The original expression becomes:
$\sin^{-1} \left( \frac{5}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right)$
We apply the addition formula for inverse sine functions: [Per the identity: $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x\sqrt{1-y^2} + y\sqrt{1-x^2} \right)$].
Here, $x = \frac{5}{13}$ and $y = \frac{3}{5}$.
Step 3: Execution of the Calculation
Substitute the values into the formula:
$\sin^{-1} \left( \left( \frac{5}{13} \right) \sqrt{1 - \left( \frac{3}{5} \right)^2} + \left( \frac{3}{5} \right) \sqrt{1 - \left( \frac{5}{13} \right)^2} \right)$
Calculate the terms within the square roots:
$\sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$
$\sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$
Compute the products:
$\sin^{-1} \left( \left( \frac{5}{13} \cdot \frac{4}{5} \right) + \left( \frac{3}{5} \cdot \frac{12}{13} \right) \right)$
$\sin^{-1} \left( \frac{20}{65} + \frac{36}{65} \right)$
$\sin^{-1} \left( \frac{56}{65} \right)$
Step 4: Geometrical Representation
The diagram illustrates the right-angled triangles corresponding to $\cos^{-1}(12/13)$ and $\sin^{-1}(3/5)$, where the sum of angles $\alpha + \beta$ results in the angle whose sine is $56/65$.
Final Solution: $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
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 $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
- 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 $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- Q11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$
- 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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