Question

If ${\cos }^4\theta \cdot {\sec }^2\alpha ,\frac{1}{2}$and ${\sin }^4\theta \cdot co\sec {}^2\alpha$ are in AP, then $\cos {}^8\theta \cdot {\sec }^6\alpha$ ,$\frac{1}{2}$ and $\sin {}^2\theta \cdot cosec{}^6\alpha$ are in

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is A AP

$\cos {}^4\theta .\sec {}^2\alpha ,\frac{1}{2}$ and $\sin {}^4\theta \csc {}^2\alpha$ are in $AP$

So,

$\cos {}^4\theta .\sec {}^2\alpha ,\frac{1}{2}and\sin {}^4\theta \csc {}^2\alpha \frac{\sin {}^4\theta \csc {}^2\alpha +\cos {}^4\theta \sec {}^2\alpha }{2}=\frac{1}{2}\sin {}^4\theta \csc {}^2\alpha +\cos {}^4\theta \sec {}^2\alpha =1or,\frac{\sin {}^4\theta }{\sin {}^2\alpha }+\frac{\cos {}^4\theta }{\cos {}^2\alpha }=1[\because \csc \theta =\frac{1}{\sin \theta },\sec \theta =\frac{1}{\cos \theta }]or,\sin {}^4\theta \cos {}^2\alpha +\cos {}^4\theta \sin {}^2\alpha =\sin {}^2\alpha or,{(1-\cos {}^2\theta )}^2\cos {}^2\alpha +\cos {}^4\theta \sin {}^2\alpha =\sin {}^2\alpha [\because 1-\sin {}^2\theta =\cos {}^2\theta ]or,(1-2\cos {}^2\theta +\cos {}^4\theta )\cos {}^2\alpha +\cos {}^4\theta \sin {}^2\alpha =\sin {}^2\alpha \cos {}^2\alpha or,\cos {}^2\alpha -2\cos {}^2\alpha \cos {}^2\alpha +\cos {}^4\theta \cos {}^2\alpha +\cos {}^4\theta \sin {}^2\alpha -\sin {}^2\alpha \cos {}^2\alpha =0or,\cos {}^2\alpha (1-\sin {}^2\alpha )-2\cos {}^2\alpha \cos {}^2\alpha +\cos {}^4\theta (\cos {}^2\alpha +\sin {}^2\alpha )=0or,\cos {}^4\alpha -2\cos {}^2\theta \cos {}^2\alpha +\cos {}^4\theta =0or,{(\cos {}^2\alpha -\cos {}^2\theta )}^2=0or,\cos {}^2\theta -\cos {}^2\alpha =0or,\cos {}^2\theta =\cos {}^2\alpha -\left(i\right)1-\sin {}^2\alpha =1-\sin {}^2\theta \sin {}^2\theta =\sin {}^2\alpha -\left(ii\right)$

Terms $\cos {}^4\theta \sec {}^2\alpha ,\frac{1}{2},\sin {}^4\theta \csc {}^2\alpha$ are in $AP$

$\cos {}^4\theta \sec {}^2\alpha ,\frac{1}{2},\sin {}^4\theta \csc {}^2\alpha \frac{\cos {}^4\theta }{\cos {}^2\alpha },\frac{1}{2},\frac{\sin {}^4\theta }{\sin {}^2\alpha }$

$\cos {}^2\theta ,\frac{1}{2},\sin {}^2\theta$ are in $AP$.

So terms,

$\cos {}^8\theta \sec {}^6\alpha ,\frac{1}{2},\sin {}^2\theta \csc {}^6\alpha \frac{\cos {}^8\theta }{\cos {}^6\theta },\frac{1}{2},\frac{\sin {}^2\theta }{\sin {}^6\theta }\cos {}^2\theta ,\frac{1}{2},\frac{1}{\sin {}^4\theta }$