Let AB be the tower.
Let C and D be the positions of the boat.
From the diagram,
\(\angle \)XAC = 60° = \(\angle \)ACB and \(\angle \) XAD = 45° = \(\angle \) ADB, BC = 200 m
In right triangle ABC, tan60° = \(\frac { AB }{ BC } \)
gives \(\sqrt { 3 } \) \(\frac { AB }{ 200 } \)
we get AB = 200\(\sqrt { 3 } \) ... (1)
In right triangle ABD, tan45° = \(\frac { AB }{ BD } \)
gives = \(\frac { 200\sqrt { 3 } }{ BD } \) [by (1)]
we get, BD = 200\(\sqrt { 3 } \)
now, CD = BD - BC
CD = 200\(\sqrt { 3 } \) - 200 = 200(\(\sqrt { 3 } \) -1) = 146.4
It is given that the distance CD is covered in 10 seconds.
That is, the distance of 146.4 m is covered in 10 seconds.
Therefore, speed of the boat = \(\frac { distance }{ time } \)
= \(\frac { 146.4 }{ 10 } \) = 14.64 m/s gives 14.64\(\times \frac { 3600 }{ 100 } \) km/hr = 52.704 km/hr

Let the observer on the lighthouse CD be at D.
Height of the lighthouse CD = 60 m
From the diagram,
\(\angle \)XDA = 28° = \(\angle \)XDB = 45°\(\angle \)DBC
In right triangle DCB,
In right triangle DCB, tan 45° = \(\frac { DC }{ BC } \)
1 = \(\frac { 60 }{ BC } \) gives BC = 60 m
In right triangle DCA, tan28° = \(\frac { DC }{ AC } \)
0.5317 = \(\frac { 60 }{ AC } \) gives AC \(\frac { 60 }{ 0.5317 } \) = 112.85
Distance between the two ships AB = AC - BC = 52.85 m

\(\frac { sinA }{ secA+tanA-1 } +\frac { cosA }{ cosecA+cotA-1 } =1\) 
\(=\frac { sinA(cosecA+cotA-1)+cosA(secA+tanA-1) }{ (secA+tanA-1)(cosecA+cotA-1) } \)
=\(\frac { sin\ A \ cosec \ A \ + \ sin \ A \ cot \ A \ - \ sin \ A\ +cos \ A \ sec \ A \ + cos \ A \ tan \ A \ - \ cos \ A }{ (sec \ A \ + \ tan \ A-1)(cosec \ A \ + \ cot \ A-1) } \)
=\(\frac { 1+cosA-sinA+1+sinA-cosA }{ \left( \frac { 1 }{ cosA } +\frac { sinA }{ cosA } -1 \right) \left( \frac { 1 }{ sinA } +\frac { cosA }{ sinA } -1 \right) } \)
=\(\frac { 2 }{ \left( \frac { 1+sinA-cosA }{ cosA } \right) \left( \frac { 1+cosA-sinA }{ sinA } \right) } \)
=\(\frac { 2sinAcosA }{ (1+sinA-cosA)(1+cosA-sinA) } \)
=\(\frac { 2 \ sin \ A \ cos \ A }{ [1+(sin \ A- \ cos \ A)][1-(sin \ A-cos \ A)] } =\frac { 2sinAcosA }{ 1-(sin \ A- \ cos \ A{ ) }^{ 2 } } \)
=\(\frac { 2sinAcosA }{ 1-(si{ n }^{ 2 }A+co{ s }^{ 2 }A-2sinAcosA) } =\frac { 2sinAcosA }{ 1-(1-2sinAcosA) } \)
=\(\frac { 2sinAcosA }{ 1-1+2sinAcosA } =\frac { 2sinAcosA }{ 2sinAcosA } =1.\)

\(\left( \frac { co{ s }^{ 3 }A-si{ n }^{ 3 }A }{ cosA-sinA } \right) -\left( \frac { co{ s }^{ 3 }A+si{ n }^{ 3 }A }{ cosA+sinA } \right) \)
\(\left( \frac { (cosA-sinA)(co{ s }^{ 2 }A+si{ n }^{ 2 }A+cosAsinA) }{ cosA-sinA } \right) \)
\(\left[ since\quad { a }^{ 3 }-{ b }^{ 3 }=(a-b)({ a }^{ 2 }+{ b }^{ 2 }+ab \ { a }^{ 3 }+{ b }^{ 3 }=(a+b)({ a }^{ 2 }+{ b }^{ 2 }-ab) \right] \)
\(\left( \frac { (cosA+sinA)(co{ s }^{ 2 }A+si{ n }^{ 2 }A-cosAsinA) }{ cosA+sinA } \right) \)
= (1 + cos A sin A) - (1 - cos A sin A)
= 2cos A sin A