A) 2
B) 1
C) \[1\frac{2}{3}\]
D) \[1\frac{1}{2}\]
Correct Answer: B
Solution :
[b] \[\sin A+{{\sin }^{2}}A=1\]\[\Rightarrow \]\[\sin A=1-{{\sin }^{2}}A\] \[\therefore \]\[\sin A={{\cos }^{2}}A\] Then, \[{{\cos }^{2}}A+{{\cos }^{4}}A={{\cos }^{2}}A+{{(co{{s}^{2}}A)}^{2}}\] \[=\sin A+{{\sin }^{2}}A=1\] |
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