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If "2sec" (2alpha) = "tan" beta + "cot"b...

If `"2sec" (2alpha) = "tan" beta + "cot"beta`, then one of the value of `alpha + beta` is

A

`(pi)/(4)`

B

`(pi)/(2)`

C

`pi`

D

none of these

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The correct Answer is:
To solve the equation \( 2 \sec(2\alpha) = \tan(\beta) + \cot(\beta) \), we can follow these steps: ### Step 1: Rewrite the given equation We start with the equation: \[ 2 \sec(2\alpha) = \tan(\beta) + \cot(\beta) \] ### Step 2: Express secant in terms of cosine Recall that \( \sec(x) = \frac{1}{\cos(x)} \). Therefore, we can rewrite the left side: \[ 2 \sec(2\alpha) = \frac{2}{\cos(2\alpha)} \] ### Step 3: Rewrite tangent and cotangent Using the definitions of tangent and cotangent: \[ \tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} \quad \text{and} \quad \cot(\beta) = \frac{\cos(\beta)}{\sin(\beta)} \] Thus, we can express the right side as: \[ \tan(\beta) + \cot(\beta) = \frac{\sin(\beta)}{\cos(\beta)} + \frac{\cos(\beta)}{\sin(\beta)} \] ### Step 4: Combine the right side into a single fraction To combine the fractions, we find a common denominator: \[ \tan(\beta) + \cot(\beta) = \frac{\sin^2(\beta) + \cos^2(\beta)}{\sin(\beta) \cos(\beta)} \] Using the Pythagorean identity \( \sin^2(\beta) + \cos^2(\beta) = 1 \), we simplify this to: \[ \tan(\beta) + \cot(\beta) = \frac{1}{\sin(\beta) \cos(\beta)} \] ### Step 5: Set the left side equal to the right side Now we have: \[ \frac{2}{\cos(2\alpha)} = \frac{1}{\sin(\beta) \cos(\beta)} \] ### Step 6: Cross-multiply Cross-multiplying gives: \[ 2 \sin(\beta) \cos(\beta) = \cos(2\alpha) \] ### Step 7: Use the double angle identity Recall that \( \sin(2\beta) = 2 \sin(\beta) \cos(\beta) \). Thus, we can rewrite the equation as: \[ \cos(2\alpha) = \sin(2\beta) \] ### Step 8: Use the complementary angle identity We know that \( \cos(x) = \sin\left(\frac{\pi}{2} - x\right) \). Therefore, we can write: \[ \cos(2\alpha) = \sin\left(\frac{\pi}{2} - 2\alpha\right) \] This implies: \[ 2\alpha = \frac{\pi}{2} - 2\beta \] ### Step 9: Rearrange the equation Rearranging gives us: \[ 2\alpha + 2\beta = \frac{\pi}{2} \] ### Step 10: Divide by 2 Dividing the entire equation by 2 results in: \[ \alpha + \beta = \frac{\pi}{4} \] ### Conclusion Thus, one of the values of \( \alpha + \beta \) is: \[ \alpha + \beta = \frac{\pi}{4} \]
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OBJECTIVE RD SHARMA ENGLISH-TRIGONOMETRIC EQUATIONS AND INEQUATIONS-Exercise
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  7. The number of points of intersection of the curves 2y =1 " and " y = "...

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  8. For m ne n, if "tan" m theta = "tan" n theta, then different values of...

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  9. If cos p theta+cos q theta=0, then prove that the different values of ...

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  10. Solutions of the equations "cos"^(2) ((1)/(2) px)+ "cos"^(2) ((1)/(2) ...

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  11. Solve sec theta-1=(sqrt(2)-1) tan theta.

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  12. If "sec"^(2) theta = sqrt(2) (1-"tan"^(2) theta), "then" theta=

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  17. The equation sin x + sin y + sin z =-3 for 0 le x le 2pi , 0 le y l...

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  19. The solution of the equation 1-"cos" theta = "sin" theta "sin" (theta)...

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  20. {x in R: "cos" 2x + 2"cos"^(2) x = 2} is equal to

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