Question

Without using trigonometric tables evaluate:-

$\frac{{\cos }^2{20}^{\circ }+{\cos }^2{70}^{\circ }}{{\sec }^2{50}^{\circ }-{\cot }^2{40}^{\circ }}+2cose{c}^2{58}^{\circ }-2\cot {58}^{\circ }\tan {32}^{\circ }-4\tan {13}^{\circ }\tan {37}^{\circ }\tan {45}^{\circ }\tan {53}^{\circ }\tan {77}^{\circ }$

• A. 1
• B. 2
• C. -1
• D. -2

Answer 1

The correct option is C -1

$\frac{{\cos }^2{20}^{\circ }+{\cos }^2{70}^{\circ }}{{\sec }^2{50}^{\circ }-{\cot }^2{40}^{\circ }}+2{\csc }^2{58}^{\circ }-2\cot {58}^{\circ }\tan {32}^{\circ }-4\tan {13}^{\circ }\tan {37}^{\circ }\tan {45}^{\circ }\tan {53}^{\circ }\tan {77}^{\circ }$

$=\frac{{\cos }^2{20}^{\circ }+{\sin }^2{20}^{\circ }}{{\sec }^2{50}^{\circ }-{\tan }^2{50}^{\circ }}+2{\csc }^2{58}^{\circ }-2\cot {58}^{\circ }\cot {58}^{\circ }-4\tan {13}^{\circ }\cot {13}^{\circ }\cot {53}^{\circ }\tan {53}^{\circ }\tan {45}^{\circ }$

$=\frac{1}{1}+2{\csc }^2{58}^{\circ }-2\cot {58}^{\circ }\cot {58}^{\circ }-4\times 1\times 1\times 1$

$=1+2({\csc }^2{58}^{\circ }-{\cot }^2{58}^{\circ })-4$

$=1+2-4$

$=-1$