Question
$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ=$ ____________.
✓
Answer
$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ=$ 1.
Solution:
$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow\Big[\frac{\cos(90-35^\circ)}{\cos55^\circ}\Big]^2+\Big[\frac{\sin(90-43)}{\sin47^\circ}\Big]^2-2\cos60^\circ$
$\Rightarrow\Big(\frac{\cos55^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\sin47^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow(1)^2+(1)^2-2\cos60^\circ\ \Big[\therefore\cos60^\circ=\frac12\Big]$
$\Rightarrow1+1-2\times\frac12$
$\Rightarrow1+1-1=1.$
Solution:
$\Big(\frac{\sin35^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\cos43^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow\Big[\frac{\cos(90-35^\circ)}{\cos55^\circ}\Big]^2+\Big[\frac{\sin(90-43)}{\sin47^\circ}\Big]^2-2\cos60^\circ$
$\Rightarrow\Big(\frac{\cos55^\circ}{\cos55^\circ}\Big)^2+\Big(\frac{\sin47^\circ}{\sin47^\circ}\Big)^2-2\cos60^\circ$
$\Rightarrow(1)^2+(1)^2-2\cos60^\circ\ \Big[\therefore\cos60^\circ=\frac12\Big]$
$\Rightarrow1+1-2\times\frac12$
$\Rightarrow1+1-1=1.$
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