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The value of cosycos(pi/2-x)-cos(pi/2-y)...

The value of `cosycos(pi/2-x)-cos(pi/2-y)cosx+sinycos(pi/2-x)+cosxsin(pi/2-y)` is zero if `x=0` (b) `y=0` `x=y` (d) `npi+y-pi/4(n in Z)`

A

`x=0`

B

`y=0`

C

`x=y`

D

`npi+y-pi/4 (n in Z)`

Text Solution

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The correct Answer is:
To solve the given expression: \[ \cos y \cos\left(\frac{\pi}{2} - x\right) - \cos\left(\frac{\pi}{2} - y\right) \cos x + \sin y \cos\left(\frac{\pi}{2} - x\right) + \cos x \sin\left(\frac{\pi}{2} - y\right) \] we will simplify it step by step. ### Step 1: Use Trigonometric Identities Recall the following trigonometric identities: - \(\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta\) - \(\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta\) Using these identities, we can rewrite the expression: \[ \cos y \sin x - \sin y \cos x + \sin y \sin x + \cos x \cos y \] ### Step 2: Group Terms Now, we can group the terms: \[ (\cos y \sin x + \sin y \sin x) + (\cos x \cos y - \sin y \cos x) \] This can be factored as: \[ \sin x (\cos y + \sin y) + \cos x (\cos y - \sin y) \] ### Step 3: Set the Expression to Zero We want to find when this expression is equal to zero: \[ \sin x (\cos y + \sin y) + \cos x (\cos y - \sin y) = 0 \] ### Step 4: Solve for Conditions This equation can be satisfied in two cases: 1. \(\sin x = 0\) or \(\cos x = 0\) 2. The coefficients must balance each other. From the first case: - If \(\sin x = 0\), then \(x = n\pi\) where \(n \in \mathbb{Z}\). - If \(\cos x = 0\), then \(x = \frac{\pi}{2} + n\pi\). From the second case, we can set up the equation: \[ \frac{\sin x}{\cos x} = \frac{\cos y - \sin y}{\cos y + \sin y} \] This leads us to the condition: \[ \tan x = \frac{\cos y - \sin y}{\cos y + \sin y} \] ### Step 5: Analyze the Conditions To find the conditions on \(y\): - If \(x = y\), then both sides are equal. - If \(y = 0\), we can find specific values of \(x\) that satisfy the equation. ### Conclusion After analyzing the conditions, we find that the expression is zero if: \[ x = n\pi + y - \frac{\pi}{4}, \quad n \in \mathbb{Z} \] Thus, the correct option is: **(d)** \(n\pi + y - \frac{\pi}{4}, \, n \in \mathbb{Z}\)

To solve the given expression: \[ \cos y \cos\left(\frac{\pi}{2} - x\right) - \cos\left(\frac{\pi}{2} - y\right) \cos x + \sin y \cos\left(\frac{\pi}{2} - x\right) + \cos x \sin\left(\frac{\pi}{2} - y\right) \] we will simplify it step by step. ...
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The value of cosycos(pi/2-x)-cos(pi/2-y)cosx+sinycos(pi/2-x)+cosxsin(pi/2-y) is zero if (A) x=0 (B) y=0 (C) x=y (D) npi+y-pi/4(n in Z)

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