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Confused! kindly explain, - Trigonometry - JEE Main-7

If \cos^{-1}x-\cos^{-1}\frac{y}{2}=\alpha , where -1\leq x\leq 1,

-2\leq y\leq 2 , x\leq \frac{y}{2}, then for all x,y,4x^{2}-4xy\cos \alpha +y^{2}

is equal to : 

  • Option 1)

    4\sin ^{2}\alpha

  • Option 2)

    2\sin ^{2}\alpha

  • Option 3)

    4\sin ^{2}\alpha-2x^{2}y^{2}

  • Option 4)

    4\cos ^{2}\alpha+2x^{2}y^{2}

 
Answers (1)
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P Plabita

\cos^{-1}x-\cos^{-1}\frac{y}{2}=\alpha                         where -1\leq x\leq 1, -2\leq y\leq 2,x\leq \frac{y}{2}

\cos \alpha =\cos (\cos^{-1}x-\cos^{-1}\frac{y}{2})

            =\cos (\cos^{-1}x)\cdot cos(\cos^{-1}\frac{y}{2})+sin(\cos^{-1}x)sin(\cos^{-1}\frac{y}{2})

            =\frac{xy}{2}+\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}

cos\alpha -\frac{xy}{2}=\sqrt{1-x^{2}}\sqrt{1-\frac{y^{2}}{4}}

Squaring both the sides

cos^{2}\alpha +\frac{x^{2}y^{2}}{4}-xycos\alpha =(1-x^{2})({1-\frac{y^{2}}{4}})

4-4x^{2}-y^{2}=4cos^{2}\alpha -4xycos\alpha

=4(1-cos^{2}\alpha )

=4sin^{2}\alpha

So, option (1) is correct.


Option 1)

4\sin ^{2}\alpha

Option 2)

2\sin ^{2}\alpha

Option 3)

4\sin ^{2}\alpha-2x^{2}y^{2}

Option 4)

4\cos ^{2}\alpha+2x^{2}y^{2}

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