Question

If tan theta =a/b, prove that asin2theta + bcos2theta=b

Answer 1

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Answer 2

Answer and Explanation:

Given : $\tan \theta =\frac{a}{b}$

To prove : $a\sin 2\theta+b\cos 2\theta=b$

Solution :

Taking LHS,

$LHS=a\sin 2\theta+b\cos 2\theta$

Using formula,

$\sin 2x=\frac{2\tan x}{1+\tan^2x}$ and $\cos 2x=\frac{1-\tan^2x}{1+\tan^2x}$

$LHS=a(\frac{2\tan \theta}{1+\tan^2\theta})+b(\frac{1-\tan^2\theta}{1+\tan^2\theta})$

Substitute, $\tan \theta =\frac{a}{b}$

$LHS=a(\frac{2(\frac{a}{b})}{1+(\frac{a}{b})^2})+b(\frac{1-(\frac{a}{b})^2}{1+(\frac{a}{b})^2})$

$LHS=a(\frac{\frac{2a}{b}}{\frac{b^2+a^2}{b^2}})+b(\frac{\frac{b^2-a^2}{b^2}}{\frac{b^2+a^2}{b^2}})$

$LHS=a(\frac{2a}{b}\times \frac{b^2}{b^2+a^2})+b(\frac{b^2-a^2}{b^2}\times\frac{b^2}{b^2+a^2})$

$LHS=\frac{2a^2b}{b^2+a^2}+\frac{b^3-a^2b}{b^2+a^2}$

$LHS=\frac{2a^2b+b^3-a^2b}{b^2+a^2}$

$LHS=\frac{a^2b+b^3}{b^2+a^2}$

$LHS=\frac{b(a^2+b^2)}{b^2+a^2}$

$LHS=b$

$LHS=RHS$

Hence proved.