| If \[\tan A+\sin \,A=p\] and \[tan\,A-\sin A=q,\] then |
A) \[{{p}^{2}}+{{q}^{2}}=4\sqrt{pq}\]
B) \[{{p}^{2}}-{{q}^{2}}=4\sqrt{pq}\]
C) \[{{p}^{2}}-{{q}^{2}}=\sqrt{pq}\]
D) \[{{p}^{2}}-{{q}^{2}}=2\sqrt{pq}\]
Correct Answer: B
Solution :
| \[\tan A+\sin A=p,\]\[\tan A-\sin A=q\] |
| \[{{(\tan A+\sin A)}^{2}}={{p}^{2}},\]\[{{(\tan A-\sin A)}^{2}}={{q}^{2}}\] |
| \[{{p}^{2}}-{{q}^{2}}=4\tan A\sin A=\frac{4{{\sin }^{2}}A}{\cos A}\] |
| \[4\sqrt{pq}=4\sqrt{({{\tan }^{2}}A-{{\sin }^{2}}A)}\] |
| \[=4\sqrt{\frac{{{\sin }^{2}}A\,(1-{{\cos }^{2}}A)}{{{\cos }^{2}}A}}=\frac{4{{\sin }^{2}}A}{\cos A}\] |
| \[\therefore \] \[{{p}^{2}}-{{q}^{2}}=4\sqrt{pq}\] |
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