Question
If $${\cos ^4}\theta \cdot {\sec ^2}\alpha ,\frac{1}{2}$$ and $${\sin ^4}\theta \cdot {\text{cose}}{{\text{c}}^2}\alpha $$ are in A.P. then $${\cos ^8}\theta \cdot {\sec ^6}\alpha ,\frac{1}{2}$$ and $${\sin ^8}\theta \cdot {\text{cose}}{{\text{c}}^6}\alpha $$ are in
A.
A.P.
B.
G.P.
C.
H.P.
D.
None of these
Answer :
A.P.
Solution :
Here, $${\cos ^4}\theta \cdot {\sec ^2}\alpha + {\sin ^4}\theta \cdot {\text{cose}}{{\text{c}}^2}\alpha = 1$$
$$\eqalign{
& {\text{or, }}{\cos ^4}\theta \cdot {\sin ^2}\alpha + {\sin ^4}\theta \cdot {\cos ^2}\alpha = {\sin ^2}\alpha \cdot {\cos ^2}\alpha \cr
& {\text{or, }}\left( {1 - {{\sin }^2}\theta } \right){\cos ^2}\theta \cdot {\sin ^2}\alpha + {\sin ^4}\theta \cdot \left( {1 - {{\sin }^2}\alpha } \right) = {\sin ^2}\alpha \left( {1 - {{\sin }^2}\alpha } \right) \cr
& {\text{or, }}{\cos ^2}\theta \,{\sin ^2}\alpha + {\sin ^4}\theta - {\sin ^2}\theta \,{\sin ^2}\alpha \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = {\sin ^2}\alpha - {\sin ^4}\alpha \cr
& {\text{or, }}{\sin ^4}\theta + {\sin ^4}\alpha - 2{\sin ^2}\theta \cdot {\sin ^2}\alpha = 0 \cr
& \Rightarrow \,\,{\sin ^2}\theta = {\sin ^2}\alpha \,\,{\text{and so, }}{\cos ^2}\theta = {\cos ^2}\alpha . \cr
& \therefore \,\,{\cos^8}\theta \cdot {\sec ^6}\alpha + {\sin ^8}\theta \cdot {\text{cose}}{{\text{c}}^6}\alpha = {\cos ^2}\theta + {\sin ^2}\theta = 1. \cr} $$