If (a^{2}-b^{2} ) \sin \\theta+2 a b \cos \\theta=a^{2}+b^{2}, then \tan \\theta=?

SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 89 of 100

Marks: +1, -0

(a^{2}-b^{2} ) \sin \θ+2 a b \cos \θ=a^{2}

+b^{2},

then

\tan \θ=?

Solution:

(2)

(a^{2}-b^{2} ) \sin \θ+2 a b \cos \θ

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

On dividing by cosθ,

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

\sec \θ

On squaring both sides,

(a^{2}-b^{2} )^{2} \tan ^{2} \θ+4 a^{2} b^{2}+4 a b

(a^{2}-b^{2} ) \tan \θ

= (a^{2}+b^{2} )^{2} \sec ^{2} \θ

⇒(a^{2}-b^{2} )^{2} \tan ^{2} \θ+4 a b (a^{2}-

b^{2} ) \tan \θ+4 a^{2} b^{2}

= (a^{2}+b^{2} )^{2} (1+\tan ^{2} \θ)

⇒(a^{2}+b^{2} )^{2} \tan ^{2} \θ- (a^{2}-b^{2} )^{2}

\tan ^{2} \θ 4 a b (a^{2}-b^{2} )^{2} \tan \θ+

(a^{2} +b^{2} )-4 a^{2} b^{2}=0

⇒\tan ^{2} \θ{ (a^{2}+b^{2} )^{2}- (a^{2}-b^{2} )^{2} }

-4 a b (a^{2}-b^{2} ) \tan \θ+ (a^{2}-b^{2} )^{2}

=0

⇒4 a^{2} b^{2} \tan ^{2} \θ-4 a b (a^{2}-b^{2} )

\tan \θ+ (a^{2}-b^{2} )^{2}=0

⇒{2 a b \tan \θ- (a^{2}-b^{2} ) }^{2}=0

⇒ 2 a b \tan \θ- (a^{2}-b^{2} )=0

⇒ \tan \θ= {a^{2}-b^{2}}/{2 a b}

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