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જો $\theta = 3\, \alpha$ અને $sin\, \theta =$ $\frac{a}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$. થાય તો $a \,cosec\, \alpha - b \,sec\, \alpha$ ની કિમત ............. થાય
$\frac{1}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$
$2 \sqrt {{a^2}\,\, + \,\,{b^2}}$
$a + b$
none
Solution
$a \,cosec\alpha – bsec\alpha $ $=$ $\frac{a}{{\sin \alpha }}\,\, – \,\,\frac{b}{{\cos \alpha }}$
$\frac{{\sqrt {{a^2} + {b^2}} }}{{\sin \alpha \,\,\cos \alpha }}\,\,\,\left[ {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\,\,\cos \alpha \, – \,\frac{b}{{\sqrt {{a^2} + {b^2}} }}\,\sin \alpha } \right]$
Now $sin3\alpha =$ $\frac{a}{{\sqrt {{a^2} + {b^2}} }}$ gives
$ \Rightarrow \,\,\sqrt {{a^2} + {b^2}} \,\,\left[ {\frac{{\sin 3\alpha \,\cos \alpha \,\, – \,\,\cos 3\alpha \,\,\sin \alpha }}{{\sin \alpha \,\,\cos \alpha }}} \right] = 2\sqrt {{a^2}\, + \,{b^2}} $
