MCQ
જો $\sin (\theta + \alpha ) = a$ અને $\sin (\theta + \beta ) = b,$ તો $\cos 2\,(\alpha - \beta ) - 4ab\,\cos (\alpha - \beta ) = . . .$
- A$1 - {a^2} - {b^2}$
- ✓$1 - 2{a^2} - 2{b^2}$
- C$2 + {a^2} + {b^2}$
- D$2 - {a^2} - {b^2}$
and $\sin \,(\theta + \beta ) = b$…..$(ii)$
Now, $\cos \,(\theta + \alpha ) = \sqrt {1 - {a^2}} \, \Rightarrow \,\,\theta + \alpha = {\cos ^{ - 1}}\sqrt {1 - {a^2}} $
and $\alpha \, - \beta = (\theta + \alpha ) - (\theta + \beta )$
$ = \,\,{\cos ^{ - 1}}\sqrt {1 - {a^2}} - {\cos ^{ - 1}}\sqrt {1 - {b^2}} $
$ \Rightarrow \,\,\alpha - \beta = {\cos ^{ - 1}}(\sqrt {1 - {a^2}} \,\sqrt {1 - {b^2}} + ab)$
$ \Rightarrow \,\,\cos \,(\alpha - \beta ) = \sqrt {1 - {a^2}} \,\sqrt {1 - {b^2}} + ab$
Now, $\cos \,\,2\,(\alpha - \beta ) - 4ab\,\,\cos \,(\alpha - \beta )$
$ = 2\,\,{\cos ^2}\,(\alpha - \beta ) - 1 - 4ab\,\,\cos \,(\alpha - \beta )$
$ = 2\,{\left( {\sqrt {1 - {a^2}} \sqrt {1 - {b^2}} + ab} \right)^2}$
$ - 4ab\,\left( {\sqrt {1 - {a^2}} \sqrt {1 - {b^2}} + ab} \right) - 1$
$ = 2\,\{ (1 - {a^2})(1 - {b^2}) + {a^2}{b^2} + 2ab\sqrt {1 - {a^2}} \sqrt {1 - {b^2}} \} $
$ - 4ab\,(\sqrt {1 - {a^2}} \sqrt {1 - {b^2}} + ab)$
$ = \,\,2\,(1 - {b^2} - {a^2} + {a^2}{b^2}) + 2{a^2}{b^2} - 4{a^2}{b^2} - 1$
$ = \,\,2\,(1 - {a^2} - {b^2}) - 1 = 1 - 2{a^2} - 2{b^2}.$
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વિધાન $- 1 :$ આ સ્પર્શકો પરસ્પર લંબ છે.
વિધાન $- 2 :$ વર્તૂળ $ x^2 + y^2 = 338$ પરના દરેક બિંદુએથી આપેલ વર્તુળ પર લંબ સ્પર્શકો દોરી શકાય.