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← JEE Advanced 2022 · Paper 2
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JEE Advanced 2022 · Paper 2, Mathematics question 1

Let α\alpha and β\beta be real numbers such that −π4<β<0<α<π4-\dfrac\pi4 < \beta < 0 < \alpha < \dfrac\pi4. If sin⁡(α+β)=13\sin(\alpha+\beta) = \dfrac13 and cos⁡(α−β)=23\cos(\alpha-\beta) = \dfrac23, then the greatest integer less than or equal to
(sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α)2\left(\frac{\sin\alpha}{\cos\beta} + \frac{\cos\beta}{\sin\alpha} + \frac{\cos\alpha}{\sin\beta} + \frac{\sin\beta}{\cos\alpha}\right)^2
is _____.

Answer and worked solution

Official answer: 1

Pair the terms.
sin⁡αcos⁡β+sin⁡βcos⁡α=sin⁡αcos⁡α+sin⁡βcos⁡βcos⁡αcos⁡β=sin⁡(α+β)cos⁡(α−β)cos⁡αcos⁡β,\frac{\sin\alpha}{\cos\beta}+\frac{\sin\beta}{\cos\alpha}=\frac{\sin\alpha\cos\alpha+\sin\beta\cos\beta}{\cos\alpha\cos\beta}=\frac{\sin(\alpha+\beta)\cos(\alpha-\beta)}{\cos\alpha\cos\beta},
cos⁡βsin⁡α+cos⁡αsin⁡β=sin⁡(α+β)cos⁡(α−β)sin⁡αsin⁡β.\frac{\cos\beta}{\sin\alpha}+\frac{\cos\alpha}{\sin\beta}=\frac{\sin(\alpha+\beta)\cos(\alpha-\beta)}{\sin\alpha\sin\beta}.

Add. With cos⁡αcos⁡β+sin⁡αsin⁡β=cos⁡(α−β)\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta):
E=sin⁡(α+β)cos⁡2(α−β)sin⁡αcos⁡αsin⁡βcos⁡β=4sin⁡(α+β)cos⁡2(α−β)sin⁡2αsin⁡2β.E=\frac{\sin(\alpha+\beta)\cos^2(\alpha-\beta)}{\sin\alpha\cos\alpha\sin\beta\cos\beta}=\frac{4\sin(\alpha+\beta)\cos^2(\alpha-\beta)}{\sin2\alpha\sin2\beta}.

The denominator.
sin⁡2αsin⁡2β=12[cos⁡2(α−β)−cos⁡2(α+β)]=12[(89−1)−(1−29)]=−49.\sin2\alpha\sin2\beta=\tfrac12\left[\cos2(\alpha-\beta)-\cos2(\alpha+\beta)\right]=\tfrac12\left[\left(\tfrac89-1\right)-\left(1-\tfrac29\right)\right]=-\tfrac49.

Value.
E=4⋅13⋅49−49=−43,E2=169≈1.78.E=\frac{4\cdot\frac13\cdot\frac49}{-\frac49}=-\frac43,\qquad E^2=\frac{16}9\approx1.78.
The greatest integer is 1.

Topics:Trigonometric ratios and identities
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