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Prove(1+cot theta -cosec theta)(1+sec theta+ tan theta)= 2

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$(1+\cot \theta - \csc \theta)(1+ \sec \theta+ \tan \theta)= 2 \\ LHS = (1+ \frac{\cos \theta }{\sin \theta } -\frac{1}{\sin \theta }) (1+\frac{1}{\cos \theta } + \frac{\sin \theta}{\cos \theta})\\ = (\frac{\sin \theta + \cos \theta -1 }{\sin \theta }) (\frac{\cos \theta +1+\sin \theta}{\cos \theta })\\ =\frac{(\sin \theta + \cos \theta)^2 -1^2 }{\sin \theta \cos \theta }\\ =\frac{\sin^2 \theta + \cos^2 \theta +2 \sin \theta \cos \theta -1^2 }{\sin \theta \cos \theta }\\ =\frac{1 +2 \sin \theta \cos \theta -1 }{\sin \theta \cos \theta }\\ =2$

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Ravindra Pindel

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