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Q4:
Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
Solution :
Step 1: Conversion of Inverse Sine Terms to Inverse Tangent Terms
To prove the identity $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$, we first express each $\sin^{-1}$ term in the form of $\tan^{-1}$ using the trigonometric identity $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.
For $\sin^{-1} \frac{8}{17}$:
- Let $\alpha = \sin^{-1} \frac{8}{17} \implies \sin \alpha = \frac{8}{17}$.
- Using the Pythagorean theorem, $\text{adj} = \sqrt{\text{hyp}^2 - \text{opp}^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$. [By Pythagorean Identity: $a^2 + b^2 = c^2$].
- Therefore, $\tan \alpha = \frac{8}{15} \implies \alpha = \tan^{-1} \frac{8}{15}$.
For $\sin^{-1} \frac{3}{5}$:
- Let $\beta = \sin^{-1} \frac{3}{5} \implies \sin \beta = \frac{3}{5}$.
- Using the Pythagorean theorem, $\text{adj} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
- Therefore, $\tan \beta = \frac{3}{4} \implies \beta = \tan^{-1} \frac{3}{4}$.
Step 2: Applying the Tangent Addition Formula
We now substitute these values into the left-hand side (LHS) of the original equation: $\tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$.
Using the inverse trigonometric formula: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$, provided $xy < 1$. [Condition check: $(\frac{8}{15}) \times (\frac{3}{4}) = \frac{24}{60} = 0.4 < 1$].
Step 3: Algebraic Computation
Substitute $x = \frac{8}{15}$ and $y = \frac{3}{4}$ into the identity:
$\text{LHS} = \tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 - (\frac{8}{15} \times \frac{3}{4})} \right)$
Calculating the numerator:
$\frac{8 \times 4 + 3 \times 15}{60} = \frac{32 + 45}{60} = \frac{77}{60}$Calculating the denominator:
$1 - \frac{24}{60} = \frac{60 - 24}{60} = \frac{36}{60}$Dividing the fractions:
$\tan^{-1} \left( \frac{77/60}{36/60} \right) = \tan^{-1} \left( \frac{77}{60} \times \frac{60}{36} \right) = \tan^{-1} \frac{77}{36}$Visual Representation of Triangle Constants
Final Solution: $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$ (Q.E.D.)
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 $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
- 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 $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- 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
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