Question

${\left(\frac{{\sin }^{m}x}{{\sin }^{n}x}\right)}^{m+n}\cdot {\left(\frac{{\sin }^{n}x}{{\sin }^{q}x}\right)}^{n+q}.{\left(\frac{{\sin }^{q}x}{{\sin }^{m}x}\right)}^{q+m}=$

• A. 2
• B. 1
• C. 3
• D. 4

Answer 1

The correct option is B 1

Consider, $({\sin }^{m-n}x{)}^{m+n}.({\sin }^{n-q}{)}^{n+q}.({\sin }^{q-m}{)}^{q+m}$

$=(\sin x{)}^{m^2-n^2}.(\sin x{)}^{n^2-q^2}.(\sin x{)}^{q^2-m^2}$

$=(\sin x{)}^{m^2+n^2+q^2-m^2-n^2-q^2}$

$=(\sin x{)}^0$

$=1$