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  • Prove that sin theta minus cos theta + 1 by sin theta + cos theta minus 1 equal to 1 by sec theta minus tan theta​

Prove that sin theta minus cos theta + 1 by sin theta + cos theta minus 1 equal to 1 by sec theta minus tan theta​

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\frac{ \sin \theta - \cos \theta +1 }{\sin \theta + \cos \theta -1 } = \frac{1}{\sec \theta -\tan \theta }\\ LHS =\frac{ \sin \theta - \cos \theta +1 }{\sin \theta + \cos \theta -1 } \\ = \frac{ \sin \theta - \cos \theta +1 }{\sin \theta + \cos \theta -1 } \times \frac{ \sin \theta + \cos \theta +1 }{\sin \theta + \cos \theta +1 }\\\\ =\frac{ (\sin \theta+1)^2 - \cos^2 \theta }{(\sin \theta + \cos \theta )^2-1^2 }\\\\ =\frac{ \sin^2 \theta + 1 + 2\sin \theta - \cos^2 \theta }{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1 }\\\\ =\frac{ 2\sin^2 \theta + 2\sin \theta }{1+ 2\sin \theta \cos \theta - 1 }\\\\ =\frac{ 2\sin \theta(\sin \theta +1 ) }{ 2\sin \theta \cos \theta }\\\\ = \frac{ \sin \theta +1 }{ \cos \theta }\\\\ =\frac{ \sin \theta +1 }{ \cos \theta } \times \frac{ 1- \sin \theta }{1- \sin \theta }\\\\ =\frac{ 1- \sin^2 \theta }{\cos \theta (1- \sin \theta) }\\\\ =\frac{ \cos \theta }{ (1- \sin \theta) }\\\\ = \frac{ 1}{sec \theta - \tan \theta } = RHS\\\\

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Ravindra Pindel

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