A) 0
B) 1
C) 2
D) -1
Correct Answer: C
Solution :
(c): \[\left( 1+sec{{40}^{{}^\circ }}-cot{{50}^{{}^\circ }} \right)\left( 1-cosec{{40}^{{}^\circ }}+tan{{50}^{{}^\circ }} \right)\] \[=\left( 1+sec{{40}^{{}^\circ }}+tan{{40}^{{}^\circ }} \right)\left( 1-cosec{{40}^{{}^\circ }}+cot{{40}^{{}^\circ }} \right)\]\[\left[ \therefore tan\left( {{90}^{{}^\circ }}-\theta \right)=cot\theta ;cot\left( {{90}^{{}^\circ }}-\theta \right)=tan\theta \right]\] \[=\left( 1+\frac{1}{\cos {{40}^{{}^\circ }}}+\frac{\sin {{40}^{{}^\circ }}}{\cos {{40}^{{}^\circ }}} \right)\] \[=\left( 1-\frac{1}{\sin {{40}^{{}^\circ }}}+\frac{\cos {{40}^{{}^\circ }}}{\sin {{40}^{{}^\circ }}} \right)\] \[=\frac{\cos {{40}^{{}^\circ }}+1+\sin {{40}^{{}^\circ }}}{\cos {{40}^{{}^\circ }}}\] \[=\frac{{{\left( \cos {{40}^{{}^\circ }}+\sin {{40}^{{}^\circ }} \right)}^{2}}-1}{\sin {{40}^{{}^\circ }}.\cos {{40}^{{}^\circ }}}\] \[=\frac{{{\cos }^{2}}{{40}^{{}^\circ }}+{{\sin }^{2}}{{40}^{{}^\circ }}+2\sin {{40}^{{}^\circ }}.\cos {{40}^{{}^\circ }}-1}{\sin {{40}^{{}^\circ }}.\cos {{40}^{{}^\circ }}}\] \[=\frac{1+2\sin {{40}^{{}^\circ }}.\cos {{40}^{{}^\circ }}-1}{\sin {{40}^{{}^\circ }}.\cos {{40}^{{}^\circ }}}=2\]
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