Question

B= tan^2 67°+ 2cos^2 16°-cotg^2 23° + 2sin^2 16°- 2cotg 37°/tan 53° A= cotg 67° . Cotg 23°-2(cos 45°. sin 64°)^2-2sin 23°/3cos 67°- sin^2 26°

Answer 1

\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)

\(=0+2\cdot1-2=0\)

\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)

\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)

\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)