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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
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The value of $$\left( {\frac{{\sin A}}{{1 - \cos A}} + \frac{{1 - \cos A}}{{\sin A}}} \right) \div \left( {\frac{{{{\cot }^2}A}}{{1 + {\text{cosec}}\,A}} + 1} \right){\text{is:}}$$

Answer & Solution
Correct Answer: Option D
$$\eqalign{ & \left( {\frac{{\sin A}}{{1 - \cos A}} + \frac{{1 - \cos A}}{{\sin A}}} \right) \div \left( {\frac{{{{\cot }^2}A}}{{1 + {\text{cosec}}\,A}} + 1} \right) \cr & = \left( {\frac{{{{\sin }^2}A + {{\left( {1 + \cos A} \right)}^2}}}{{\sin A\left( {1 - \cos A} \right)}}} \right) \div \left( {\frac{{\frac{{{{\cos }^2}A}}{{{{\sin }^2}A}} \times \sin A}}{{1 + \sin A}} + 1} \right) \cr & = \frac{{{{\sin }^2}A + 1 + {{\cos }^2}A - 2\cos A}}{{\sin A\left( {1 - \cos A} \right)}} \div \frac{{{{\cos }^2}A + \sin A + {{\sin }^2}A}}{{\sin A\left( {1 + \sin A} \right)}} \cr & = \frac{{2\left( {1 - \cos A} \right)}}{{\sin A\left( {1 - \cos A} \right)}} \times \frac{{\sin A\left( {1 + \sin A} \right)}}{{\left( {1 + \sin A} \right)}} \cr & = 2 \cr} $$
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