Question

lf ${\cos }^4\theta {\sec }^2\alpha ,\frac{1}{2}$ and ${\sin }^4\theta cose{c}^2\alpha$ are in $\text{A.P.}$, then ${\cos }^8\theta {\sec }^6\alpha ,\frac{1}{2}$ and ${\sin }^8\theta cose{c}^6\alpha$ are in

• A. A.P
• B. G.P
• C. H.P
• D. none of these

Answer 1

Answer: (A)

A.P

We have:
${\cos }^4\theta .{\sec }^2\alpha +{\sin }^4\theta .co{\sec }^2\alpha =1=>\frac{{(1-{\sin }^2\theta )}^2}{1-{\sin }^2\alpha }+\frac{{\sin }^4\theta }{{\sin }^2\alpha }=1=>{\sin }^2\alpha +{\sin }^4\theta {\sin }^2\alpha -2{\sin }^2\theta {\sin }^2\alpha +{\sin }^4\theta -{\sin }^4\theta {\sin }^2\alpha ={\sin }^2\alpha -{\sin }^4\alpha =>{\sin }^4\theta -2{\sin }^2\theta {\sin }^2\alpha +{\sin }^4\alpha =0=>{({\sin }^2\theta -{\sin }^2\alpha )}^2=0=>{\sin }^2\alpha ={\sin }^2\theta$
Thus:
${\cos }^8\theta {\sec }^6\alpha +{\sin }^8\theta {cosec}^6\alpha ={\cos }^8\theta {\sec }^6\theta +{\sin }^8\theta {cosec}^6\theta ={\cos }^2\theta +{sin}^2\theta =1$
Hence, option A is correct.