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 = \(69.4°\)

\(\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta \)
\(\text { LHS }=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \)
\( =\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{(1+\sin \theta)^{2}}{1^{2}-\sin ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{1^{2}-\sin ^{2} \theta}} \)
\(=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}} \)
\(=\frac{1+\sin \theta}{\cos \theta}+\frac{1-\sin \theta}{\cos \theta} \)
\(=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta} \)
\(=2 \times \frac{1}{\cos \theta} \)
\(=2 \sec \theta \)
= RHS

\( \frac{\cos \theta}{1+\sin \theta} =\sec \theta-\tan \theta \)
\(\text { LHS } =\frac{\cos \theta}{1+\sin \theta} \)
\(=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta} \)
[Multiplying the Numerator and Denominator by 1 - sin\(\theta\)]
\( =\frac{\cos \theta(1-\sin \theta)}{1^{2}-\sin ^{2} \theta} \)
\(\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta} \)
\(\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{(1-\sin \theta)}{\cos \theta} \)
\(=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta-\tan \theta \)
= RHS

tan⁴\(\theta \) + tan²\(\theta \) = sec⁴\(\theta \) - sec²\(\theta \) 
L.H.S = tan²θ (tan²θ + 1)
= tan²θ.sec²θ 
= sec⁴θ - sec²θ
= R.H.S

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

Let BC be the height of the tower and A be the position of the ball lying on the ground. Then,
BC = 20 m and \(\angle\)XCA = 60° = \(\angle\)CAB
Let AB = x metres.
In the right angled ΔABC,
tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 } =\frac { 20 }{ x } \)
\(x=\frac { 20\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\frac { 20\times 1.732 }{ 3 } \) = 11.54m
Hence, the distance between the foot of the tower and the ball is 11.55 m.

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.

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

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