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If y=(sin^(4)x-cos^(4)x+sin^(2) x cos^(2...

If `y=(sin^(4)x-cos^(4)x+sin^(2) x cos^(2)x)/(sin^(4) x+ cos^(4)x + sin^(2) x cos^(2)x), x in (0, pi/2)`, then

A

`-(3)/(2) le y le 1/2`

B

` 1 le y le 1/2`

C

`-(5)/(3) le y le 1`

D

None of these

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The correct Answer is:
To solve the given problem, we need to simplify the expression for \( y \): \[ y = \frac{\sin^4 x - \cos^4 x + \sin^2 x \cos^2 x}{\sin^4 x + \cos^4 x + \sin^2 x \cos^2 x} \] ### Step 1: Use the identity for \( a^4 - b^4 \) Recall that \( a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) \). We can apply this to \( \sin^4 x - \cos^4 x \): \[ \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) \] Since \( \sin^2 x + \cos^2 x = 1 \), we have: \[ \sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x \] ### Step 2: Substitute back into the expression for \( y \) Now substitute this back into the expression for \( y \): \[ y = \frac{\sin^2 x - \cos^2 x + \sin^2 x \cos^2 x}{\sin^4 x + \cos^4 x + \sin^2 x \cos^2 x} \] ### Step 3: Simplify the denominator Next, we simplify the denominator \( \sin^4 x + \cos^4 x \). We can use the identity: \[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \] Thus, the denominator becomes: \[ \sin^4 x + \cos^4 x + \sin^2 x \cos^2 x = (1 - 2\sin^2 x \cos^2 x) + \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x \] ### Step 4: Substitute the denominator back into \( y \) Now substitute this back into the expression for \( y \): \[ y = \frac{\sin^2 x - \cos^2 x + \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} \] ### Step 5: Factor the numerator The numerator can be rearranged: \[ y = \frac{(\sin^2 x - \cos^2 x) + \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} \] ### Step 6: Final simplification At this point, we can evaluate \( y \) for specific values of \( x \) in the interval \( (0, \frac{\pi}{2}) \) or analyze its behavior. However, we can also see that: \[ \sin^2 x - \cos^2 x = \sin^2 x - (1 - \sin^2 x) = 2\sin^2 x - 1 \] Thus, we can write: \[ y = \frac{(2\sin^2 x - 1) + \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} \] This expression can be evaluated further or analyzed depending on the context of the problem.
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ARIHANT MATHS ENGLISH-TRIGONOMETRIC FUNCTIONS AND IDENTITIES-Exercise (Questions Asked In Previous 13 Years Exam)
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  7. For 0 lt theta lt pi/2 , the solution (s) of sum(m=1)^6cos e c(theta+(...

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  10. cos(alpha-beta)=1a n dcos(alpha+beta)=l/e , where alpha,betamu in [-pi...

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  15. If A = sin^2x + cos^4 x, then for all real x :

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  16. Let cos(alpha+beta)""=4/5 and let sin (alpha+beta)""=5/(13) where 0lt=...

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  19. If 0 lt x lt pi and cos x + sin x = 1/2, then tan x is

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