class 12 maths determinants

Evaluate $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$.

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📘 Determinants NCERT,Misc.Q.3,Page.141 SA

Evaluate $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$.

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha }\\{ - \sin \beta }&{\cos \beta }&0\\{\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right|$

Expanding along ${R_1},$

we get
$= \cos \alpha \cos \beta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cos \beta }&0\\{\sin \alpha \sin \beta }&{\cos \alpha }\end{array}} \right| - \cos \alpha \cos \beta \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \beta }&0\\{\sin \alpha \cos \beta }&{\cos \alpha }\end{array}} \right| - \sin \alpha \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - \sin \beta }&{\cos \beta }\\{\sin \alpha \sin \beta }&{\sin \alpha \sin \beta }\end{array}} \right|$

$= \cos \alpha \cos \beta (\cos \beta \cos \alpha ) - \cos \alpha \sin \beta ( - \sin \beta \cos \alpha ) - \sin \alpha ( - \sin \alpha {\sin ^2}\beta - \sin \alpha {\cos ^2}\beta )$

$= {\cos ^2}\alpha {\cos ^2}\beta + {\cos ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\sin ^2}\beta + {\sin ^2}\alpha {\cos ^2}\beta$

$= {\cos ^2}\alpha \cdot 1 + {\sin ^2}\alpha \cdot 1 = {\cos ^2}\alpha + {\sin ^2}\alpha = 1.$

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