Question

The value of $\frac{\sin \theta \cos \theta .\sin ({90}^o-\theta )}{\cos ({90}^o-\theta )}+\frac{\cos \theta .\sin \theta .\cos ({90}^o-\theta )}{\sin ({90}^o-\theta )}+\frac{{\sin }^2{27}^o+{\sin }^2{63}^o}{{\cos }^2{40}^o+{\cos }^2{50}^o}$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is B $2$

$\sin ({90}^o-\theta )=\cos \theta ;\cos ({90}^o-\theta )=\sin \theta$
$\sin {27}^o=\sin ({90}^o-{63}^o)={\cos 63}^o$
${\cos 40}^o=\cos ({90}^o-{50}^o)=\sin {50}^o$
$\therefore Givenexp.$
$=\frac{\sin \theta .\cos \theta .\cos \theta }{\sin \theta }+\frac{\cos \theta .\sin \theta \sin \theta }{\cos \theta }+\frac{({\cos }^2{63}^o+{\sin }^2{63}^o)}{({\sin }^2{50}^o+{\cos }^2{50}^o)}$
$={\cos }^2\theta +{\sin }^2\theta +\frac{1}{1}=1+1=2$