Question

please solve this question ​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Given :

$a = sin \alpha cos \beta$

$b = sin \alpha sin \beta$

To Find :

${a}^{2} + {b}^{2} + {c}^{2}$

$= > {(sin \alpha cos \beta )}^{2} + {(sin \alpha sin \beta )}^{2} + {(co s\alpha )}^{2}$

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

Taking sin alpha common :

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

$\bold{\boxed{\red{{sin \beta }^{2} + {cos}^{2} \beta = 1}}}$

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

$\bold{1\: is\: your \:answer}$

Other Formulas:

• sin (90-theta)=cos theta
• Cos(90-theta)=sin theta