A) 1
B) 2
C) 4
D) None of these
Correct Answer: B
Solution :
[b] We have, \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=\pi -{{\sin }^{-1}}z\] or \[x\sqrt{(1-{{y}^{2}})}+y\sqrt{(1-{{x}^{2}})}=z\] or \[{{x}^{2}}(1-{{y}^{2}})={{z}^{2}}+{{y}^{2}}(1-{{x}^{2}})-2yz\sqrt{(1-{{x}^{2}})}\] or \[{{({{x}^{2}}-{{z}^{2}}-{{y}^{2}})}^{2}}=4{{y}^{2}}{{z}^{2}}(1-{{x}^{2}})\] or \[{{x}^{4}}+{{y}^{4}}+{{z}^{4}}-2{{x}^{2}}{{z}^{2}}+2{{y}^{2}}{{z}^{2}}-2{{x}^{2}}{{y}^{2}}\] \[+4{{x}^{2}}{{y}^{2}}{{z}^{2}}-4{{y}^{2}}{{z}^{2}}=0\] or \[{{x}^{4}}+{{y}^{4}}+{{z}^{4}}+4{{x}^{2}}{{y}^{2}}{{z}^{2}}=2({{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{z}^{2}}{{x}^{2}})\] \[\therefore k=2\]You need to login to perform this action.
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