Question

Evaluate: $\frac{\sin \theta \cos \theta \sin ({90}^o-\theta )}{\cos ({90}^o-\theta )}+\frac{\cos \theta \sin \theta \cos ({90}^o-\theta )}{\sin ({90}^o-\theta )}+\frac{{\sin }^2{27}^o+{\sin }^2{63}^o}{{\cos }^2{40}^o+{\cos }^2{50}^o}$

• A. $1$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is B $2$

$\frac{\sin \theta \cos \theta \sin 90-\theta )}{\cos (90-\theta )}+\frac{\cos \theta \sin \theta \cos (90-\theta )}{\sin 90-\theta )}+\frac{{\sin }^{2}27+{\sin }^{2}63}{{\cos }^{2}40+{\cos }^{2}50}\frac{\sin \theta \cos \theta \sin 90-\theta )}{\cos 90-\theta )}=\frac{\sin \theta \cos \theta \cos \theta }{\sin \theta }={\cos }^2\theta ..\left(1\right)[\because \sin 90-\theta )=\cos \theta ]\frac{\cos \theta \sin \theta \cos (90-\theta )}{\sin 90-\theta )}=\frac{\cos \theta \sin \theta \sin \theta }{\cos \theta }={\sin }^2\theta ....(2\left)\right[\because \cos 90-\theta )=\sin \theta ]\frac{{\sin }^{2}27+{\sin }^2(90-27)}{{\cos }^{2}40+{\cos }^2(90-40)}=\frac{{\sin }^{2}27+{\cos }^{2}27}{{\cos }^{2}40+{\sin }^{2}40}(As{\sin }^2\theta +{\cos }^2\theta =1)=\frac{1}{1}=1$

Using (1),(2),(3), we get

${\sin }^2\theta +{\cos }^2\theta +1=1+1=2$