| If \[A+B=90{}^\circ ,\] then the value of \[\frac{2\,\,({{\sin }^{2}}A+{{\sin }^{2}}B)}{\text{cose}{{\text{c}}^{2}}\,\,(A+B)}\]is |
A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[2\]
D) \[1\]
Correct Answer: C
Solution :
| \[A+B=90{}^\circ \]\[\Rightarrow \]\[A=(90{}^\circ -B)\] |
| \[\therefore \]\[\frac{2\,\,[{{\sin }^{2}}A+{{\sin }^{2}}(90{}^\circ -A)]}{\text{cose}{{\text{c}}^{2}}(90{}^\circ )}=\frac{2\,\,({{\sin }^{2}}A+{{\cos }^{2}}A)}{\text{cose}{{\text{c}}^{2}}90{}^\circ }\] |
| \[=\frac{2}{1}=2\] |
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