A) (a) \[\cos \theta \,sin\theta -\frac{\sin \theta \cos ({{90}^{o}}-\theta )\cos \theta }{\sec ({{90}^{o}}-\theta )}\] \[-\frac{\cos \theta sin({{90}^{o}}-\theta )\sin \theta }{\cos ec({{90}^{o}}-\theta )}=0\] (b) If A and B are complementary angles, then \[\sin A=\sqrt{\frac{\cos A}{\sin B}-\cos A\,\sin B}\] Only (a)
B) Only (b)
C) Neither (a) nor (b)
D) Both (a) and (b)
Correct Answer: D
Solution :
(a) \[\cos \theta \,\sin \theta -\frac{\sin \theta \cos ({{90}^{o}}-\theta )\,\cos \theta }{\sec ({{90}^{o}}-\theta )}\] \[-\frac{\cos \theta sin({{90}^{o}}-\theta )\,\sin \theta }{\cos ec({{90}^{o}}-\theta )}\] \[=\cos \theta \,\sin \theta -{{\sin }^{3}}\theta \,\cos \theta -co{{s}^{3}}\theta \,\sin \theta \] \[=\cos \theta \,\sin \theta -\cos \theta \,\sin \theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[=\cos \theta \,\sin \theta -\cos \theta \,\sin \theta =0\] (b) A and B are complementary angles \[\Rightarrow \] \[A+B={{90}^{o}}\] \[\Rightarrow \] \[A={{90}^{o}}-B\] Now, taking R.H.S we get \[\sqrt{\frac{\cos A}{\sin B}-\cos A\sin B}\] \[=\sqrt{\frac{\cos ({{90}^{o}}-B)}{\sin B}-\cos ({{90}^{o}}-B)\sin B}\] \[=\sqrt{\frac{\sin B}{\sin B}-\sin B\sin B}\] \[=\sqrt{1-{{\sin }^{2}}B}=\sqrt{{{\cos }^{2}}B}\] \[=\cos B=\cos ({{90}^{o}}-A)=\sin A\]
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