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If 2 cos ( A//2) = sqrt(( 1+ sin A )) - ...

If `2 cos ( A//2) = sqrt(( 1+ sin A )) - sqrt(( 1- sin A )) `, then

A

`npi + ( pi //4) lt A//2 lt npi + ( 3pi //4)`

B

`npi - ( pi //4) lt A//2 lt 2npi - ( 3pi //4)`

C

`2npi - ( 3pi //4) lt A//2 lt 2npi + ( 5pi //4)`

D

`2npi + ( pi //4) lt A//2 lt 2npi + ( 3pi //4)`

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AI Generated Solution

The correct Answer is:
To solve the equation \( 2 \cos \left( \frac{A}{2} \right) = \sqrt{1 + \sin A} - \sqrt{1 - \sin A} \), we can follow these steps: ### Step 1: Square both sides Start by squaring both sides of the equation to eliminate the square roots. \[ \left(2 \cos \left( \frac{A}{2}\right)\right)^2 = \left(\sqrt{1 + \sin A} - \sqrt{1 - \sin A}\right)^2 \] ### Step 2: Simplify the left-hand side The left-hand side simplifies to: \[ 4 \cos^2 \left( \frac{A}{2} \right) \] ### Step 3: Expand the right-hand side Using the identity \((a - b)^2 = a^2 - 2ab + b^2\), we expand the right-hand side: \[ \left(\sqrt{1 + \sin A}\right)^2 - 2 \sqrt{1 + \sin A} \sqrt{1 - \sin A} + \left(\sqrt{1 - \sin A}\right)^2 \] This simplifies to: \[ (1 + \sin A) + (1 - \sin A) - 2 \sqrt{(1 + \sin A)(1 - \sin A)} \] ### Step 4: Combine like terms Combining the terms gives: \[ 2 - 2 \sqrt{1 - \sin^2 A} \] Using the Pythagorean identity \(1 - \sin^2 A = \cos^2 A\), we can rewrite this as: \[ 2 - 2 \cos A \] ### Step 5: Set the equation Now we have: \[ 4 \cos^2 \left( \frac{A}{2} \right) = 2 - 2 \cos A \] ### Step 6: Divide by 2 Dividing both sides by 2 gives: \[ 2 \cos^2 \left( \frac{A}{2} \right) = 1 - \cos A \] ### Step 7: Use the double angle formula Recall the double angle formula for cosine: \[ \cos A = 2 \cos^2 \left( \frac{A}{2} \right) - 1 \] Substituting this into our equation gives: \[ 2 \cos^2 \left( \frac{A}{2} \right) = 1 - (2 \cos^2 \left( \frac{A}{2} \right) - 1) \] ### Step 8: Simplify the equation This simplifies to: \[ 2 \cos^2 \left( \frac{A}{2} \right) = 2 - 2 \cos^2 \left( \frac{A}{2} \right) \] ### Step 9: Combine like terms Adding \(2 \cos^2 \left( \frac{A}{2} \right)\) to both sides gives: \[ 4 \cos^2 \left( \frac{A}{2} \right) = 2 \] ### Step 10: Solve for \(\cos^2 \left( \frac{A}{2} \right)\) Dividing both sides by 4 results in: \[ \cos^2 \left( \frac{A}{2} \right) = \frac{1}{2} \] ### Step 11: Find \(\cos \left( \frac{A}{2} \right)\) Taking the square root gives: \[ \cos \left( \frac{A}{2} \right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \] ### Step 12: Determine possible angles The angles for which \(\cos \left( \frac{A}{2} \right) = \frac{\sqrt{2}}{2}\) are: \[ \frac{A}{2} = 45^\circ + n \cdot 360^\circ \quad \text{or} \quad \frac{A}{2} = 315^\circ + n \cdot 360^\circ \] Thus, multiplying by 2 gives: \[ A = 90^\circ + n \cdot 720^\circ \quad \text{or} \quad A = 630^\circ + n \cdot 720^\circ \]
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ML KHANNA-TRIGONOMETRY RATIOS AND IDENTITIES-MISCELLANEOUS EXERCISE (SELF ASSESSMENT TEST)
  1. If tan theta = 3 and theta lies in thrid quadrant, then the value of s...

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  2. If sin alpha = 12//13 , ( 0 lt alpha lt pi //2) and cos beta = - ( 3)...

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  3. If theta and phi are angles in the first quadrant such that tan theta ...

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  4. The value of cos ( pi //5) cos ( 2pi //5) cos ( 4pi //5) cos ( 8pi //5...

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  5. sin 20^(@) sin 40^(@) sin 60^(@) sin 80^(@) =(3)/(16)

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  6. The value of sqrt3 cosec20^@-sec20^@ is equal to:

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  7. Givent that pi/2,alphaltpi then the expression sqrt((1-sinalpha)/(1+si...

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  8. If tan alpha =(b)/(a), a gt b gt 0 and " if " 0 lt alpha lt (pi)/(4), ...

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  9. If cos(A-B)=3/5and tan A tanB=2, then

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  10. If A +B = ( pi )/( 3) and cos A + cos B =1, then which of the followin...

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  11. If Sigma n^2 = lambda Sigma n, then sin^(-1) "" ((9 lambda^2 - 4 n^2)/...

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  12. If A = 580^(@) then

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  13. If 2 cos ( A//2) = sqrt(( 1+ sin A )) - sqrt(( 1- sin A )) , then

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  14. If cos(alpha+beta)=4/5, sin(alpha-beta)=5/13and alpha, beta between 0 ...

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  15. If a Delta PQR " if" 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P =1 , ...

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  16. If (sin^4x)/2+(cos^4x)/3=1/5t h e n tan^2x=2/3 (b) (sin^8x)/8+(cos^...

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  17. Let theta , phi in [0,2pi ] be such that 2 cos theta ( 1- sin phi ) = ...

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  18. The maximum value of the expression 1/(sin^2theta+3sinthetacostheta+5c...

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  19. The expression (tanA)/(1-cotA)+(cotA)/(1-tanA) can be written as :

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  20. ABCD is a trapezium such that AB and CD are parallel and BC bot CD. If...

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