Question

Without using trigonometric tables find the value of

$\frac{2}{3}\left(\frac{\sec {56}^{\circ }}{cosec{34}^{\circ }}\right)-2{\cos }^2{20}^{\circ }+\frac{1}{2}\cot {28}^{\circ }\cot {35}^{\circ }\cot {45}^{\circ }\cot {62}^{\circ }\cot {55}^{\circ }-2{\cos }^2{70}^{\circ }$

• A. $\frac{4}{5}$
• B. $-\frac{3}{4}$
• C. $-\frac{5}{6}$
• D. 1

Answer 1

Answer: (C)

$-\frac{5}{6}$

$\frac{2}{3}\left(\frac{\sec 56°}{cosec34°}\right)-2{\cos }^{2}20°+\frac{1}{2}\cot 28°\cot 45°\cot 62°\cot 55°-2{\cos }^{2}70°$

$=\frac{2}{3}\left(\frac{\sec \left(90°34°\right)}{cosec34°}\right)-2{\cos }^2\left(90°70°\right)-2{\cos }^{2}70°+\frac{1}{2}\cot (90°-62°).1.\cot 62°$

$=\frac{2}{3}\left(\frac{cosec34°}{cosec34°}\right)-2\sin 70°-2{\cos }^{2}70°+\frac{1}{2}\tan 62°.\cot 62°$

$=\frac{2}{3}-2({\sin }^{2}70°+{\cos }^{2}70°)+\frac{1}{2}$

$=\frac{2}{3}-2+\frac{1}{2}$

$=\frac{4-12+3}{6}$

$=\frac{-5}{6}$

Hence, the answer is $\frac{-5}{6}.$