Let AD is the first building.
BC is the second building.
AD = BE = BC - CE
o
From the right triangle CED
\(\tan 45^{\circ} =\frac{C E}{D E} \)
\(1 =\frac{C E}{A B}=\frac{C E}{70 m} \)
CE = 70 m
BE = BC - EC
= 120 m - 70 m = 50 m
AD = 50 m
Height of the first building is 50 m.

Let AB be the rock.
C be the position of the car.
In right triangle ABC
\(\tan 30^{\circ} =\frac{A B}{B C} \)
\(\frac{1}{\sqrt{3}} =\frac{50 \sqrt{3}}{B C} \)
\(B C =50 \sqrt{3} \times \sqrt{3} \)
= 50 x 3 = 150 m
Distance of the car from the rock = 150 m.

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

From the right \(\triangle\)ABC
\( \tan \theta =\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{A C}{B C} \)
\(=\frac{10 \sqrt{3} m}{30 m}=\frac{\sqrt{3}}{3} \)
\(=\frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\tan \theta =\frac{1}{\sqrt{3}} \)
\(\theta =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
0 = 30^o
Angle of elevation is 30^o

Let PQ the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.in right triangle PQR,\(\angle \)PRQ=30°
tan\(\theta =\frac { PQ }{ QR } \)
tan30° = \(\frac { h }{ 48 } \) gives,\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 48 } \) so, h =\(16\sqrt { 3 } \)
Therefore the height of the tower \(16\sqrt { 3 } \) 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 = \(38.7°\)

\( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} =\sec \theta+\tan \theta \)
\(\mathbf{L H S} =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}\)
[Multiplying the Numerator and denominator by \(\sqrt{1-\sin \theta}\)]
\( =\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\sqrt{\frac{\cos ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{\cos \theta}{1-\sin \theta} \)
\(=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
[Multiplying Numerator and denominator by \(1+\sin \theta\)]
\( =\frac{\cos \theta(1+\sin \theta)}{1^{2}-\sin ^{2} \theta}=\frac{\cos \theta(1+\sin \theta)}{\cos ^{2} \theta} \)
\({\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \)
\(=\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta+\tan \theta=\text { RHS }\)

\( \frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} =\tan ^{2} \theta \)
\(\text { LHS } =\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1} \)
\(=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1} \)
\(=\frac{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta}}{\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}\)
\(=\frac{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\left(\cos ^{2} \theta-\sin ^{2} \theta\right)} \)
\(=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\tan ^{2} \theta=\text { RHS }\)

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

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

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

sec\(\theta \) - cos\(\theta \)  = \(\frac { 1 }{ cos\theta } -cos\theta =\frac { 1-co{ s }^{ 2 }\theta }{ cos\theta } \)
= \(\frac { si{ n }^{ 2 }\theta }{ cos\theta } \) [since 1 - cos²\(\theta \) = sin²\(\theta \)]
= \(\frac { sin\theta }{ cos\theta } \times sin\theta =tan\theta sin\theta \) 

1 + \(\frac { co{ t }^{ 2 }\theta }{ 1+cosec\theta +1 } \) = 1+ \(\frac { cose{ c }^{ 2 }\theta -1 }{ cosec\theta +1 } \) [since cosec²-1 = cot²\(\theta \)
= 1+\(\frac { (cosec\theta +1)(cosec\theta -1) }{ cosec\theta +1 } \) 
1 +( cosec\(\theta \)-1) = cosec\(\theta \)

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

tan² \(\theta \) - sin ²\(\theta \) = tan² \(\theta \) -\(\frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \),cos²\(\theta \) 
= tan² \(\theta \) (1-cos² \(\theta \) ) = tan² \(\theta \) sin^{2 \(\theta \)}