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AP and TS EAMCET Engineering Model Paper 20

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Question : 20 of 160

Marks: +1, -0

If

\cos (\theta-\alpha)= a

\cos (\theta-\beta)= b,

then the value of

\sin^{2}(\alpha-\beta)+2 a b \cos(\alpha-\beta)

$a^{2}-b^{2}$
$a^{2}+b^{2}$
$-(a^{2}+b^{2})$
$b^{2}-a^{2}$

Solution:

Here

\sin(\alpha-\beta)=a \sin(\theta-\beta)-b \sin(\theta-\gamma)

\sin^{2}(\alpha-\beta)=a^{2} \sin^{2}(\theta-\beta)

+b^{2} \sin^{2}(\theta-\alpha)-2 a b \sin(\theta-\beta) \sin(\theta-\alpha)

\cos(\alpha-\beta)=\cos(\theta-\beta) \cos(\theta-\alpha)

+\sin(\theta-\beta) \sin(\theta-\alpha)

=a b+\sin(\theta-\beta) \sin(\theta-\alpha)

2 a b \cos(\alpha-\beta)=2 a^{2} b^{2}

+2 a b \sin(\theta-\beta) \sin(\theta-\alpha)

\sin^{2}(\alpha-\beta)+2 a b \cos(\alpha-\beta)

=a^{2} (1-b^{2})+b^{2} (1-a^{2})+2 a^{2} b^{2}

=a^{2}+b^{2}

Here

(\alpha-\beta)=(\theta-\beta)-(\theta-\alpha)

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