in right triangle ABC [see fig.(b)]
tan\(\theta \) =\(\frac { 8 }{ 3 } \)
= tan^-1(2.66)
\(\theta \) = \(69.4°\)(since tan \(69.4°\)=2.6604)
\(\angle \)BAC = \(69.4°\)

The height of the first building AB = 60 m. Now, AB = MD = 60 m
Let the height of the second building CD = h. Distance BD = 140 m
Now, AM = BD = 140 m
From the diagram,
\(\angle \)XCA = 30° =\(\angle \)CAM
In right triangle AMC, tan30° = \(\frac { CM }{ Am } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { CM }{ 140 } \)
CM=\(\frac { 140 }{ \sqrt { 3 } } =\frac { 140\sqrt { 3 } }{ 3 } \)
\(=\frac { 140\times 1.732 }{ 3 } \)
CM = 80.78
Now, h = CD = CM + MD = 80.78 + 60 = 140.78
Therefore the height of the second building is 140.78 m

in the right triangle ABC [see figure. (a)]
tan \(\theta \) =\(\frac { opposite\ side\ }{ adjacent\ side\ } =\frac { 4 }{ 5 } \)
= tan^-1(0.8)
\(\theta \) = \(38.7°\)(since tan \(38.7°\) = 0.8011)
\(\angle \)BAC = \(38.7°\)

\( \cot \theta+\tan \theta=\sec \theta \operatorname{cosec} \theta \)
\(\text { LHS } =\cot \theta+\tan \theta \)
\(=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta} \)
\(=\frac{1}{\sin \theta} \times \frac{1}{\cos \theta} \)
\(=\operatorname{cosec} \theta \sec \theta \)
\(=\sec \theta \operatorname{cosec} \theta \)
= RHS
LHS = RHS

\(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } = \frac { \frac { 1 }{ cos\theta } }{ sin\theta } -\frac { sin\theta }{ cos\theta } =\frac { 1 }{ sin\theta cos\theta } -\frac { sin\theta }{ cos\theta } \)
\(=\frac { 1-si{ n }^{ 2 }\theta }{ sin\theta cos\theta } =cot\theta \)

\(\frac { sinA }{ 1+cosA } = \)\(\frac { sinA }{ 1+cosA } \)\(\times \frac { 1-cosA }{ 1-cosA } \) [ multiply numerator and denominator by the conjugate of 1+cosA]
= \(\frac { sinA(1-cosA) }{ (1+cosA)\quad (1-cosA) } =\frac { sinA(1-cosA) }{ 1-co{ s }^{ 2 }A } \) 
= \(\frac { sinA(1-cosA) }{ si{ n }^{ 2 }A } =\frac { 1-cosA }{ sinA } \) 

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

\(\mathrm{LHS}=\cot ^{2} A\left(\frac{\sec A-1}{1+\sin A}\right)+\sec ^{2} A\left(\frac{\sin A-1}{1+\sec A}\right)\)
\( =\frac{\cot ^{2} A(\sec A-1)(\sec A+1)+\sec ^{2} A(\sin A-1)(1+\sin A)}{(1+\sin A)(1+\sec A)} \)
\(=\frac{\cot ^{2} A\left(\sec ^{2} A-1\right)+\sec ^{2} A\left(\sin ^{2} A-1\right)}{(1+\sin A)(1+\sec A)} \)
\(=\frac{\cot ^{2} A \tan ^{2} A-\sec ^{2} A\left(1-\sin ^{2} A\right)}{(1+\sin A)(1+\sec A)} \)
\( =\frac{\cot ^{2} A \tan ^{2} A-\sec ^{2} A \cos ^{2} A}{(1+\sin A)(1+\sec A)} \)
\(=\frac{\frac{1}{\tan ^{2} A} \tan ^{2} A-\frac{1}{\cos ^{2} A} \cos ^{2} A}{(1+\sin A)(1+\sec A)} \)
\(=\frac{1-1}{(1+\sin A)(1-\sec A)}=0=\text { RHS } \)

Let AC be the height of the tower.
Let AB be the height of the building.
Then, AC = h metres, AB = 30m
In right triangle CBP,\(\angle \)CPB = \(60°\)
tan\( \theta \) = \(\frac { BC }{ BP } \)
tan \(60°\) \(\frac { AB+AC }{ BP } so,\sqrt { 3 } =\frac { 30+h }{ BP } \)      ...(1)
In right triangle ABP,\(\angle \)APB = \(45°\) 
\(tan\theta =\frac { AB }{ BP } \) 
tan\(45°\)=\(\frac { 30 }{ BP } \) gives BP = 30        ....(2)
substituting (2) in (1). We get \(\sqrt { 3 } =\frac { 30+h }{ 30 } \) 
h = 30\((\sqrt { 3 } -1)\) = 30(1.732 - 1) = 30(0.732) = 21.96m.
Hence, the height of the tower is 21.96 m.

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