| (i) If \[x=a{{\cos }^{3}}\theta ,\,y=b\,{{\sin }^{3}}\theta \] then\[{{\left( \frac{x}{a} \right)}^{2/3}}+{{\left( \frac{y}{b} \right)}^{2/3}}=\underline{P}.\] |
| (ii) If \[x=a\sec \theta \,\,cos\phi ,\] \[y=b\,\sec \theta \sin \phi \] and \[z=c\,\tan \theta ,\] then \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{z}^{2}}}{{{c}^{2}}}=\underline{Q}.\] |
| (iii) If \[\cos A+{{\cos }^{2}}a=1,\] then \[{{\sin }^{2}}A+{{\sin }^{4}}A=\underline{R}.\] |
A)
P Q R 1 3 1
B)
P Q R 4 1 2
C)
P Q R 2 2 1
D)
P Q R 1 1 1
Correct Answer: D
Solution :
(i) We have, \[x=a{{\cos }^{2}}\theta \] and \[y=b{{\sin }^{3}}\theta \] \[\therefore \] \[\frac{x}{a}={{\cos }^{3}}\theta \] and \[\frac{y}{b}={{\sin }^{3}}\theta \] \[{{\left( \frac{x}{a} \right)}^{2/3}}={{\cos }^{2}}\theta \] and\[{{\left( \frac{y}{b} \right)}^{2/3}}={{\sin }^{2}}\theta \] Hence, \[{{\left( \frac{x}{a} \right)}^{2/3}}+{{\left( \frac{y}{b} \right)}^{2/3}}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \] \[P=1\]. (ii) We have, \[x=a\,\sec \theta \cos \phi ,\] \[y=b\,\sec \theta \sin \phi \] and \[z=c\,\tan \theta \] \[\therefore \] \[\frac{x}{a}=\sec \theta \cos \phi ,\frac{y}{b}=\sec \theta \sin \phi ,\frac{z}{c}=\tan \theta \] Hence, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-\frac{{{z}^{2}}}{{{c}^{2}}}={{(\sec \theta \,cos\phi )}^{2}}+{{(\sec \theta \sin \phi )}^{2}}\]\[-{{(\tan \theta )}^{2}}\] \[={{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1+{{\tan }^{2}}\theta -{{\tan }^{2}}\theta =1\] \[\therefore \] \[Q=1.\] (iii) \[\cos A+{{\cos }^{2}}A=1\] (Given) ?..(i) \[\therefore \] \[\cos A=1-{{\cos }^{2}}A={{\sin }^{2}}A\] \[\therefore \] \[{{\sin }^{2}}A+{{\sin }^{4}}A=\cos A+{{\cos }^{2}}A=1\] \[\therefore \] \[R=1\]
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