Question

$\frac{\cot (90°-\mathrm{\theta })·\sin (90°-\mathrm{\theta })}{\mathrm{sin\theta }}+\frac{\cot 40°}{\tan 50°}-({\cos }^{2}20°+{\cos }^{2}70°)=?$
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

$$\begin{gathered} (c)\; 1 \\ \mathrm{W}\text{e have:} \\ [\frac{\cot ({90}^0-\theta ).\sin ({90}^0-\theta )}{\sin \theta }+\frac{\cot {40}^0}{\tan {50}^0}-({\cos }^2{20}^0+{\cos }^2{70}^0)] \\ =\left[\frac{\tan \theta .\cos \theta }{\sin \theta }+\frac{\cot ({90}^0-{50}^0)}{\tan {50}^0}-\{{\cos }^2({90}^0-{70}^0)+{\cos }^2{70}^0\}\right]\left[\begin{array}{l}\because \cot ({90}^0-\theta )=\tan \theta \text{and}\\ \sin ({90}^0-\theta )=\cos \theta \end{array}\right] \\ =[\frac{\frac{\sin \theta }{\sout{c}\sout{o}\sout{s}\sout{\theta }}.\sout{c}\sout{o}\sout{s}\sout{\theta }}{\sin \theta }+\frac{\tan {50}^0}{\tan {50}^0}-({\sin }^2{70}^0+{\cos }^2{70}^0)][\because \text{cos}({90}^0-\theta )\text{=}\sin \theta ] \\ =(\frac{\sin \theta }{\sin \theta }+1-1) \\ =1+1-1=1 \end{gathered}$$