SSC CGL Tier-2 01-December-2016 Maths Question 89

Question 89

If $$(a^2 - b^2)\sin \theta + 2ab \cos \theta = a^2 + b^2$$, then $$\tan \theta =$$

Solution

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

divide it by $$ a^2 + b^2$$

we get

$$\frac{(a^2 - b^2)\sin \theta}{a^2 + b^2} + \frac{2ab \cos \theta}{a^2 + b^2} = 1$$ ($$\because \sin^2 \theta + \cos^2 \theta = 1$$)

here $$\sin \theta = \frac{(a^2 - b^2)}{a^2 + b^2} $$

$$\cos \theta = \frac{2ab}{a^2 + b^2} $$

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$= $$ \frac{(a^2 - b^2)}{2ab}$$



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