If (a²-b²)θ+2abθ=a²+b², then θ=?

Correct Answer is : a2−b22ab{a²-b²}{2ab}2aba2−b2

Correct Answer is : a2−b22ab{a²-b²}{2ab}2aba2−b2

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SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 89 of 100

Marks: +1, -0

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

+b^{2},

\tan\theta=?

Solution:

(2)

(a^{2}-b^{2})\sin\theta+2ab\cos\theta

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

On dividing by cosθ,

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

\sec\theta

On squaring both sides,

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

(a^{2}-b^{2})\tan\theta

=(a^{2}+b^{2})^{2}\sec^{2}\theta

\Rightarrow (a^{2}-b^{2})^{2}\tan^{2}\theta+4ab(a^{2}-

b^{2})\tan\theta+4a^{2}b^{2}

=(a^{2}+b^{2})^{2}(1+\tan^{2}\theta)

\Rightarrow (a^{2}+b^{2})^{2}\tan^{2}\theta-(a^{2}-b^{2})^{2}

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

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

\Rightarrow \tan^{2}\theta\left((a^{2}+b^{2})^{2}-(a^{2}-b^{2})^{2}\right)

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

=0

\Rightarrow 4a^{2}b^{2}\tan^{2}\theta-4ab(a^{2}-b^{2})

\tan\theta+(a^{2}-b^{2})^{2}=0

\Rightarrow \left(2ab\tan\theta-(a^{2}-b^{2})\right)^{2}=0

\Rightarrow 2ab\tan\theta-(a^{2}-b^{2})=0

\Rightarrow \tan\theta = \frac{a^{2}-b^{2}}{2ab}

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