A) sec2a
B) cosec2a+ sec2\[\beta \]
C) cosec2 \[\alpha \] cosec2 \[\beta \]
D) sec2 \[\alpha \] cosec2 \[\beta \]
Correct Answer: C
Solution :
\[cose{{c}^{2\text{ }}}\alpha +cose{{c}^{2}}\text{ }\beta \] \[=\frac{1}{{{\sin }^{2}}\alpha }+\frac{1}{{{\sin }^{2}}\beta }\] \[=\frac{{{\sin }^{2}}\beta +{{\sin }^{2}}a}{{{\sin }^{2}}\alpha .{{\sin }^{2}}\beta }\] \[=\frac{{{\cos }^{2}}\alpha +{{\sin }^{2}}a}{{{\sin }^{2}}\alpha .{{\sin }^{2}}\beta }\] \[[\because \alpha +\beta =90{}^\circ \Rightarrow \sin \,\beta =\sin (90{}^\circ -\alpha )=\cos \alpha ]\] \[=\frac{1}{{{\sin }^{2}}\alpha .{{\sin }^{2}}\beta }\] \[={{\operatorname{cosec}}^{2}}\alpha .\cos e{{c}^{2}}\beta .\]
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