Question

The value of ${\sin }^{2}A{\cos }^{2}B+{\cos }^{2}A{\cos }^{2}B+{\sin }^{2}A{\sin }^{2}B$ is...

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $1$

As we know that ${\sin }^2\theta +{\cos }^2\theta =1{\sin }^{2}A.{\cos }^{2}B+{\cos }^{2}A{\sin }^{2}B+{\cos }^{2}A.{\cos }^{2}B+{\sin }^{2}A.{\sin }^{2}B={\sin }^{2}A({\cos }^{2}A+{\sin }^{2}B)+{\cos }^{2}A({\sin }^{2}B+{\cos }^{2}B)\Rightarrow {\sin }^{2}A\left(1\right)+{\cos }^{2}A\left(1\right)\Rightarrow {\sin }^{2}A+{\cos }^{2}A=1$

Option A is the correct answer.