Question

Question number 11 please from the given attachment.

Answer 1

Hope u get the answer

Ans is 1

Answer 2

your question is ----->
$\frac{11}{7}\frac{sin70}{cos20}-\frac{4}{7}\frac{cos53.cosec37}{tan15.tan35.tan55.tan75}$

we know,
$\qquad sin(90^{\circ}-A)=cosA\\\qquad sin70^{\circ}=sin(90^{\circ}-20^{\circ})=cos20^{\circ}$

again, $tan(90-A)=cotA$
tan15° = tan(90° - 75°) = cot75°
tan35° = tan(90° - 55°) = cot55°

$cossec(90- A) = secA$
cosec37° = cosec(90° - 53°) = sec53°

now,
$\frac{11}{7}\frac{cos20}{cos20}-\frac{4}{7}\frac{cos53.sec53}{cot15.cot55.tan55.tan75}\\\\=\frac{11}{7}-\frac{4}{7}\\\\=\frac{11-4}{7}=1$

hence, answer is 1