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If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P ,

then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6 \alpha $ are in A.P

Now i have reached upto

$1=\cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha$. By completing square i have $(\sin^2 \theta \cot\alpha - \cos^2 \theta \tan \alpha)^2=0$ so i get $\tan \theta= \pm \tan \alpha$ How do i proceed?

Thanks

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We have $\cos^4\theta\sec^2\alpha+\sin^4\theta\csc^2\alpha=1$

Set $\cos^2\theta=a,\sin^4\theta=(1-a)^2$ to solve for $a$ to find $a=\cos^2\alpha$

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  • $\begingroup$ We can use $$\sec^2\alpha+\csc^2\alpha=\sec^2\alpha\csc^2\alpha$$ $\endgroup$ Feb 13 '17 at 11:13
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You have done rightly. Now we can proceed as follows: $$\tan \theta = \pm \tan \alpha \Rightarrow \tan^2 \theta = \tan^2 \alpha \Rightarrow \sec^2 \theta -1 =\sec^2 \alpha -1 \Rightarrow \sec^2 \theta = \sec^2 \alpha \Leftrightarrow \sec^6 \theta = \sec^6 \alpha \tag{1}$$

Also, $$\tan^2 \theta = \tan^2 \alpha \Rightarrow \cot^2 \theta = \cot^2 \alpha \Rightarrow \csc^2 \theta -1 = \csc^2 \alpha -1 \Rightarrow \csc^2 \theta = \csc^2 \alpha \Leftrightarrow \csc^6 \theta = \csc^6 \alpha \tag {2}$$

Hope you can take it from here by using $(1)$ and $(2)$.

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  • $\begingroup$ i got it thanks a lot $\endgroup$
    – Taylor Ted
    Feb 13 '17 at 10:52

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