Evaluate $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$.
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$
Expanding along ${R_1},$
we get
$= \cos \alpha \cos \beta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \beta }&0\\{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right| - \cos \alpha \cos \beta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \beta }&0\\{\sin \alpha \cos \beta }&{\cos \alpha }\end{array}} \right| - \sin \alpha \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \beta }&{\cos \beta }\\{\sin \alpha \sin \beta }&{\sin \alpha \sin \beta }\end{array}} \right|$
$= \cos \alpha \cos \beta (\cos \beta \cos \alpha ) - \cos \alpha \sin \beta ( - \sin \beta \cos \alpha ) - \sin \alpha ( - \sin \alpha {\sin ^2}\beta - \sin \alpha {\cos ^2}\beta )$
$= {\cos ^2}\alpha {\cos ^2}\beta + {\cos ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\cos ^2}\beta$
$= {\cos ^2}\alpha \cdot 1 + {\sin ^2}\alpha \cdot 1 = {\cos ^2}\alpha + {\sin ^2}\alpha = 1.$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Determinants. Curated by Sachin Sharma. Free for all students.