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, \[sin\text{ }C\text{ }cos\text{ }C\text{ }tan\text{ }C+co{{s}^{2}}\,C\text{ }sin\text{ }A\] \[=\frac{\sqrt{x}}{\sqrt{x}+5}\frac{\sqrt{5}}{\sqrt{x}+5}\frac{\sqrt{x}}{\sqrt{5}}+{{\left( \frac{\sqrt{5}}{\sqrt{x}+5} \right)}^{2}}.1\] \[=\frac{x}{x+5}+\frac{5}{x+5}\] \[=\frac{x+5}{x+5}\] \[=1\]
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