Solving Inverse Trigonometric Expression
The question asks for the value of the expression \(\cot^2 (\sec^{-1}2) + \tan^2 (\text{cosec}^{-1}3)\). This involves evaluating terms with inverse trigonometric functions and then summing them.
Step-by-Step Calculation
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Evaluate \(\cot^2 (\sec^{-1}2)\)
- Let \(A = \sec^{-1}2\).
- By definition of inverse secant, this means \(\sec A = 2\).
- We know that \(\sec A = \frac{\text{hypotenuse}}{\text{adjacent}}\) in a right-angled triangle. We can assume a triangle where the hypotenuse is 2 and the adjacent side is 1.
- Using the Pythagorean theorem (\(opposite^2 + adjacent^2 = hypotenuse^2\)), we find the opposite side: \(opposite^2 + 1^2 = 2^2 \implies opposite^2 = 4 - 1 = 3 \implies opposite = \sqrt{3}\).
- Now, we need \(\cot A\). We know \(\cot A = \frac{\text{adjacent}}{\text{opposite}}\).
- So, \(\cot A = \frac{1}{\sqrt{3}}\).
- Therefore, \(\cot^2 (\sec^{-1}2) = \cot^2 A = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\).
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Evaluate \(\tan^2 (\text{cosec}^{-1}3)\)
- Let \(B = \text{cosec}^{-1}3\).
- By definition of inverse cosecant, this means \(\text{cosec} B = 3\).
- We know that \(\text{cosec} B = \frac{\text{hypotenuse}}{\text{opposite}}\). We can assume a triangle where the hypotenuse is 3 and the opposite side is 1.
- Using the Pythagorean theorem (\(opposite^2 + adjacent^2 = hypotenuse^2\)), we find the adjacent side: \(1^2 + adjacent^2 = 3^2 \implies adjacent^2 = 9 - 1 = 8 \implies adjacent = \sqrt{8} = 2\sqrt{2}\).
- Now, we need \(\tan B\). We know \(\tan B = \frac{\text{opposite}}{\text{adjacent}}\).
- So, \(\tan B = \frac{1}{2\sqrt{2}}\).
- Therefore, \(\tan^2 (\text{cosec}^{-1}3) = \tan^2 B = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}\).
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Add the results
- The expression is \(\cot^2 (\sec^{-1}2) + \tan^2 (\text{cosec}^{-1}3)\).
- Substituting the calculated values: \(\frac{1}{3} + \frac{1}{8}\).
- To add these fractions, find a common denominator, which is \(3 \times 8 = 24\).
- \(\frac{1}{3} + \frac{1}{8} = \frac{1 \times 8}{3 \times 8} + \frac{1 \times 3}{8 \times 3} = \frac{8}{24} + \frac{3}{24}\).
- Adding the numerators: \(\frac{8 + 3}{24} = \frac{11}{24}\).
Final Answer
The value of the expression \(\cot^2 (\sec^{-1}2) + \tan^2 (\text{cosec}^{-1}3)\) is \(\frac{11}{24}\).
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