If \tan \theta=\frac{b}{a}, then the value of a \cos 2 \theta+b \sin 2 \theta is

NIMCET 2011 Question Paper with solutions

Question : 3 of 120

Marks: +1, -0

\tan θ=\;{b}/{a}

, then the value of

a \cos 2 θ+b \sin 2 θ

Solution:

Given,

\tan θ=\;{b}/{a}

∴ \;\; a \cos 2 θ+b \sin 2 θ

=a(\;{1-\tan ^{2} θ}/{1+\tan ^{2} θ})+b(\;{2 \tan θ}/{1+\tan ^{2} θ})

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

=a(\;{a^{2}-b^{2}}/{a^{2}+b^{2}})+b(\;{2ab}/{a^{2}+b^{2}})

=\;{a(a^{2}+b^{2})}/{a^{2}+b^{2}}=a

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