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AOB is a diameter of a circle with centre O and C is any point on the circle, joining A, C, B, C, and O, C, prove that `cosec^(2)angleCAB-1=tan^(2)angleABC`. |
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Answer» since AOB is a diameter and C is any point on the circle, `:.angleACB` is a semicircular angle. `:.angleACB=90^(@)` `:.` AB is a hypotenuse of the right-angled triangle ABC. Again, since `angleACB=90^(@),:.angleBAC+angleCBA=90^(@)`. `cosec^(2)angleCAB-1` `=cot^(2)angleCAB` `=cot^(2)(90^(@)-angleABC)` `=tan^(2)angleABC`. Hence `cosec^(2)angleCAB-1=tan^(2)angleABC`. |
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