question_answer

In the \[\Delta \,ABC\](see figure), \[\angle A=\] right angle, \[AB=\sqrt{x}\] and \[BC=\sqrt{x+5}\]. Evaluate

\[sin\text{ }C.\text{ }cos\text{ }C.\text{ }tan\text{ }C+co{{s}^{2}}C.\text{ }sin\text{ }A\]

Answer:

In \[\Delta \text{ }ABC\], by pythagoras theorem.

\[{{\left( \sqrt{x+5} \right)}^{2}}={{\left( \sqrt{x} \right)}^{2}}+A{{C}^{2}}\]

\[x+5=x+A{{C}^{2}}\]

\[5=A{{C}^{2}}\]

or                               \[AC=\sqrt{5}\]

\[\sin \,C=\frac{\sqrt{x}}{\sqrt{x}+5};\cos \,C=\frac{\sqrt{5}}{\sqrt{x}+5};\]

\[\tan \,\,C=\frac{\sqrt{x}}{\sqrt{5}}\]

and                           \[\sin \,A=\sin \,90{}^\circ \]

\[=1\]

Then, \[sin\text{ }C\text{ }cos\text{ }C\text{ }tan\text{ }C+co{{s}^{2}}\,C\text{ }sin\text{ }A\]

\[=\frac{\sqrt{x}}{\sqrt{x}+5}\frac{\sqrt{5}}{\sqrt{x}+5}\frac{\sqrt{x}}{\sqrt{5}}+{{\left( \frac{\sqrt{5}}{\sqrt{x}+5} \right)}^{2}}.1\]

\[=\frac{x}{x+5}+\frac{5}{x+5}\]

\[=\frac{x+5}{x+5}\]

\[=1\]