Question

Prove that (cosec A - sin A) (sec A - cos A ) = 1/tanA + cot A
A boat goes 24 km upstream and 28 km down stream in 6 hrs . It goes 30 km upstream and 21 km downstream in 6 h 30m. Find the speed of boat in still water.

Answer 1

1)
[tex](\csc{A}-\sin{A})(\sec{A}-\cos{A})=(\frac1{\sin{A}}-\sin{A})(\frac1{\cos{A}}-\cos{A}})=\\\frac{1-\sin^2{A}}{\sin{A}}\cdot\frac{1-\cos^2{A}}{\cos{A}}=\frac{\cos^2{A}\sin^2{A}}{\sin{A}\cos{A}}=\sin{A}\cos{A}\\\\\frac1{\tan{A}}+\cot{A}=\cot{A}+\cot{A}=2\cot{A}=\frac{2\cos{A}}{\sin{A}}\ne\sin{A}\cos{A}\\\\\textrm{because when:}\\\\\frac{2\cos{A}}{\sin{A}}=\sin{A}\cos{A}\\\\\textrm{For}\ \sin{A}\ne0\textrm{ and}\ \cos{A}\ne0:\\2\cos{A}=\sin^2{A}\cos{A}\\\sin^2{A}=2\\\sin{A}=\pm\sqrt2\approx\pm1.4\\\textrm{but:}[/tex]

[tex]\\\forall{A\in{R}:}\ \ -1\le\sin{A}\le1\\\textrm{The equation is false, so:}\\(\csc{A}-\sin{A})(\sec{A}-\cos{A})\ne\frac1{\tan{A}}+\cot{A}[/tex]

2)
$v_b=\textrm{boat speed}\\v_w=\textrm{water speed}\\v_u=v_b-v_w=\textrm{upstream speed}\\v_d=v_b+v_w=\textrm{downstream speed}\\\\time=\frac{distance}{speed}\\\\\frac{24}{v_u}+\frac{28}{v_d}=6\\\\\frac{30}{v_u}+\frac{21}{v_d}=6.5\\\\\frac{24v_d+28v_u}{v_uv_d}=6\\\\\frac{30v_d+21v_u}{v_uv_d}=6.5\\\\24v_d+28v_u}=6v_uv_d\\30v_d+21v_u=6.5v_uv_d\\\\6v_uv_d-28v_u=24v_d\\v_u(6v_d-28)=24v_d\\v_u=\frac{24v_d}{6v_d-28}=\frac{12v_d}{3v_d-14}\\\\30v_d+21\cdot\frac{12v_d}{3v_d-14}=6.5v_d\cdot\frac{12v_d}{3v_d-14}\\30+\frac{21\cdot12}{3v_d-14}=6.5\cdot\frac{12v_d}{3v_d-14}\\30\cdot(3v_d-14)+21\cdot12=6.5\cdot12v_d\\90v_d-78v_d=168\\12v_d=168\\v_d=14\\v_u=\frac{12v_d}{3v_d-14}=\frac{12\cdot14}{3\cdot14-14}=\frac{12\cdot14}{2\cdot14}=6\\\\v_b+v_w=14\\v_b-v_w=6\\-------\\2v_b=20\\v_b=10\textrm{ km/h}\v_w=14-10=4\textrm{ km/h}$

Answer:
Speed of boat in still water is 10 km/h.