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If $\, a\sin^2\theta+b\cos^2\theta=m$, $\, a\cos^2\phi+b\sin^2\phi=n$, $\, a\tan\theta=b\tan\phi$. Prove that

$$\frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}$$

I've tried setting : $$ \frac{a}{b}=\frac{\tan\theta}{\tan\phi}=k \\ a=bk \\ \tan\theta=k\tan\phi $$

and substituting these in the equations. Can anyone tell me what I'm doing wrong ? A shorter path to the answer would also be appreciated.

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$$a\sin^2\theta+b\cos^2\theta=m\to a\tan^2 \theta+b=m\sec^2\theta\\ a\tan^2\theta +b=m(1+\tan^2\theta)\to \tan^2\theta=\frac{m-b}{a-m}$$

so, using the same idea

$$\tan^2\phi=\frac{n-a}{b-n}$$

we also have that

$$\frac{\tan^2 \theta}{\tan^2 \phi}=\frac{b^2}{a^2}$$

Can you finish?

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  • $\begingroup$ Yeah. Got it. Thanks. $\endgroup$ – Vishnu V.S Mar 17 '17 at 15:19

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