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Evaluate : 3cos80°.cosec10° + 2cos59°.cosec31°
Answer & Solution
Correct Answer:
Option
D
3cos80°.cosec10° + 2cos59°.cosec31°
[If A + B = 90° then, cosA.cosecB = 1]
(3 × 1 + 2 × 1 = 5)
(the value of sinθ is always between -1 to +1)
$$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$
$$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$ $${\text{2cos5}}{{\text{9}}^ \circ }$$ $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$
$$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$
[If A + B = 90° then, cosA.cosecB = 1]
(3 × 1 + 2 × 1 = 5)
(the value of sinθ is always between -1 to +1)
$$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin {{10}^ \circ }}}{\text{ + 2cos5}}{{\text{9}}^ \circ }\frac{1}{{{\text{sin 3}}{{\text{1}}^ \circ }}}$$
$$ \Rightarrow {\text{3cos8}}{{\text{0}}^ \circ }\frac{1}{{\sin \left( {{{90}^ \circ } - {{80}^ \circ }} \right)}}{\text{ + }}$$ $${\text{2cos5}}{{\text{9}}^ \circ }$$ $$\frac{1}{{{\text{sin }}{{\left( {{{90}^ \circ } - {{59}^ \circ }} \right)}^ \circ }}}$$
$$\eqalign{ & \Rightarrow 3 + 2 \cr & \Rightarrow 5 \cr} $$
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