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If sin^(4)A-cos^(4)A=1,then (A//2)is .(0...

If `sin^(4)A-cos^(4)A=1,then (A//2)`is ______.`(0ltAle90^(@))`.

A

`45^(@)`

B

`60^(@)`

C

`30^(@)`

D

`40^(@)`

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The correct Answer is:
To solve the equation \( \sin^4 A - \cos^4 A = 1 \), we can follow these steps: ### Step 1: Recognize the difference of squares The expression \( \sin^4 A - \cos^4 A \) can be factored using the difference of squares formula: \[ \sin^4 A - \cos^4 A = (\sin^2 A + \cos^2 A)(\sin^2 A - \cos^2 A) \] Since \( \sin^2 A + \cos^2 A = 1 \), we can simplify the equation to: \[ 1(\sin^2 A - \cos^2 A) = 1 \] Thus, we have: \[ \sin^2 A - \cos^2 A = 1 \] ### Step 2: Rewrite \( \sin^2 A \) From the equation \( \sin^2 A - \cos^2 A = 1 \), we can express \( \sin^2 A \) in terms of \( \cos^2 A \): \[ \sin^2 A = \cos^2 A + 1 \] ### Step 3: Use the Pythagorean identity Using the identity \( \sin^2 A + \cos^2 A = 1 \), we can substitute for \( \sin^2 A \): \[ \cos^2 A + 1 + \cos^2 A = 1 \] This simplifies to: \[ 2\cos^2 A + 1 = 1 \] ### Step 4: Solve for \( \cos^2 A \) Rearranging gives: \[ 2\cos^2 A = 0 \] Thus: \[ \cos^2 A = 0 \] ### Step 5: Find \( A \) Since \( \cos^2 A = 0 \), we have: \[ \cos A = 0 \] The angle \( A \) that satisfies this in the range \( 0 < A < 90^\circ \) is: \[ A = 90^\circ \] ### Step 6: Calculate \( \frac{A}{2} \) Now, we find \( \frac{A}{2} \): \[ \frac{A}{2} = \frac{90^\circ}{2} = 45^\circ \] ### Final Answer Thus, the value of \( \frac{A}{2} \) is \( 45^\circ \). ---

To solve the equation \( \sin^4 A - \cos^4 A = 1 \), we can follow these steps: ### Step 1: Recognize the difference of squares The expression \( \sin^4 A - \cos^4 A \) can be factored using the difference of squares formula: \[ \sin^4 A - \cos^4 A = (\sin^2 A + \cos^2 A)(\sin^2 A - \cos^2 A) \] Since \( \sin^2 A + \cos^2 A = 1 \), we can simplify the equation to: ...
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PEARSON IIT JEE FOUNDATION-TRIGONOMETRY-Level 1
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  8. (tan^(3)theta-1)/(tantheta-1)= .

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