question_answer

ABCD is a rectangle of which AC is a diagonal. The value of \[(\tan {{\,}^{2}}\angle CAD+1)\,si{{n}^{2}}\angle BAC\]is

A) \[2\]

B) \[\frac{1}{4}\]

C) \[1\]

D) \[0\]

Correct Answer: C

Solution :

[c] According to the question, \[\because \]       ABCD is a rectangle \[\therefore \]      \[\angle \,DAC=\frac{90{}^\circ }{2}=45{}^\circ \] \[\therefore \]      \[(ta{{n}^{2}}\,\angle \,CAD+1)\,si{{n}^{2}}\,\angle \,BAC\] \[=(ta{{n}^{2}}45{}^\circ +1)\ si{{n}^{2}}45{}^\circ \] \[={{\left( 1+1\times \frac{1}{\sqrt{2}} \right)}^{2}}\] \[=2\times \frac{1}{2}=1\]