Let C be the point where the ball is
ㄥC = 60° (alternate angles)
In ΔABC, \(tan\ 60^0={AB\over BC}\)
⇒ \(\sqrt3={20\over x}\)
⇒ \(x={20\over \sqrt3}\)
\(=20\left(\sqrt3\over3 \right)\)
Hence, required distance is
1 = 11.53m

Let height of flagstaff = x m
\(tan\ 30^0={AB\over AP}\)
⇒  \({1\over \sqrt3}={10\over AP}\)
AP=10√3
i.e., distance of the building
= 10 x 1.732= 17.32m

\(tan\ 45^0={AD\over Ap}\ or\ 1={10+x\over 10\sqrt3}\)
or length of flagstaff x = 10(√3 -1) = 7.32m.

Let AB be the lighthouse and C be the position of the boat
Since ㄥPAC = 60. ㄥACB = 60°
Let CB = x
In ΔABC, \({AB\over Bc}=tan\ 60^0\)
\(\Rightarrow\ \ {40\over x}=\sqrt3\)
\(\Rightarrow\ \ x={40\over \sqrt3}\times{\sqrt3 \over\sqrt3}\)
\(={40\sqrt3\over3}m\)
The boat is \({40\sqrt3\over3}m\) away from the foot of lighthouse.

\(2\cot ^{ 2 }{ A } -1=2\left( { cosec }^{ 2 }A-1 \right) -1\)
\(=\frac { 2 }{ \sin ^{ 2 }{ A } } -3\)
\(\left( \because \quad \cot ^{ 2 }{ \theta } =-1+{ cosec }^{ 2 }\theta \right) \)
\(=\frac { 2 }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } } -3\)
\(2\cot ^{ 2 }{ A } -1=\frac { 8 }{ 3 } -3=\frac { -1 }{ 3 } \)

\(LHS=\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } } =\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } \times \frac { 1-\cos { A } }{ 1-\cos { A } } } \)
\(=\sqrt { \frac { { \left( 1-\cos { A } \right) }^{ 2 } }{ \left( 1-\cos ^{ 2 }{ A } \right) } } =\sqrt { \frac { { \left( 1-\cos { A } \right) }^{ 2 } }{ \sin ^{ 2 }{ A } } } \)
\(\left( \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right) \)
\(=\frac { 1-\cos { A } }{ \sin { A } } =\frac { 1 }{ \sin { A } } -\frac { \cos { A } }{ \sin { A } } \)
\(=cosecA-\cot { A } =RHS\)

\(LHS=-1+\frac { \sin { A } \sin { \left( { 90 }^{ ° }-A \right) } }{ \cot { \left( { 90 }^{ ° }-A \right) } } \)
\(\left[ \because \quad \sin { \left( { 90 }^{ ° }-\theta \right) =\cos { \theta } } \right] \)
\(\left[ \because \quad \cot { \left( { 90 }^{ ° }-\theta \right) } =\tan { \theta } \right] \)
\(=-1+\frac { \sin { A } \cos { A } }{ \tan { A } } \)
= - 1 + sin A cos A x cot A
\(\left[ \because \quad \cot { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \right] \)
\(\left[ \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=-1+\sin { A } \cos { A } \times \frac { \cos { A } }{ \sin { A } } \)
= - 1 + cos² A = - (1 - cos²A)
= - sin²A = RHS

Given : 4 cos \(\theta\)  = 11 sin   \(\theta\)
\(\Rightarrow \quad \cos { \theta } =\frac { 11 }{ 4 } \sin { \theta } \)
Now \(\frac { 11\cos { \theta } -7\sin { \theta } }{ 11\cos { \theta } +7\sin { \theta } } =\frac { 11\times \frac { 11 }{ 4 } \sin { \theta } -7\sin { \theta } }{ 11\times \frac { 11 }{ 4 } \sin { \theta } +7\sin { \theta } } \)
\(=\frac { \sin { \theta } \left( \frac { 121 }{ 4 } -7 \right) }{ \sin { \theta } \left( \frac { 121 }{ 4 } +7 \right) } \)
\(=\frac { 121-28 }{ 121+28 } =\frac { 93 }{ 149 } \)

Given, \(\sqrt { 2 } \sin { \theta } =1\)
\(\sin { \theta } =\frac { 1 }{ \sqrt { 2 } } =\sin { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \) = \(\sec ^{ 2 }{ { 45 }^{ ° } } -cosec^{ 2 }{ 45 }^{ ° }\)
\(={ \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 }\)
= 2 - 2
= 0 

\(\frac { \sin { { 90 }^{ ° } } }{ \cos { { 45 }^{ ° } } } +\frac { 1 }{ cosec{ 30 }^{ ° } } =\frac { 1 }{ \frac { 1 }{ \sqrt { 2 } } } +\frac { 1 }{ 2 } \)
\(=\sqrt { 2 } +\frac { 1 }{ 2 } \)
\(=\frac { 2\sqrt { 2 } +1 }{ 2 } \)

Given, \(\angle CBA=90^0,\angle BRS=90^0,\angle BSC=90^0,\) BC = 7.5 cm, RS = 5 cm, BR = 6 cm and AB = 18 cm
Then, AR = AB - RB = 18 - 6 = 12 cm

In \(\triangle ABC,\)  by using Pythagoras theorem, we get  
AS² = AR² + RS² = (12)² + (5)²
AS² = 144 + 25 = 169
\(\Rightarrow\) AS = \(\sqrt 169\)
\(\therefore\)  AS = 13 cm [since, side cannot be negative]
\((i)\quad In\quad \triangle BRS,\ tan{ x }^{ 0 }=\frac { P }{ B } =\frac { BR }{ RS } =\frac { 6 }{ 5 } \quad \)
\((ii)\quad In\quad \triangle ARS,\quad sin{ y }^{ 0 }=\frac { P }{ H } =\frac { SR }{ AS } =\frac { 5 }{ 13 } \)
\((iii)\quad In\quad \triangle ARS,\quad cos{ y }^{ 0 }=\frac { AR }{ AS } =\frac { 12 }{ 13 } \)

So, the given equation is rewritten as
\(\sqrt{3} x^{2}-4 x+\sqrt{3}=0\)
Using quadratic formula, we get
\(x=\sqrt{3}, \frac{1}{\sqrt{3}} \text { i.e. } \cot \theta=\sqrt{3}, \frac{1}{\sqrt{3}}\) \(\text { of, } \tan \theta=\frac{1}{\sqrt{3}}, \sqrt{3} \text { . }\)
\(\frac{10}{3}\) 

\(\frac { \sqrt { 2+\sqrt { 3 } } }{ 2 } \) 

Given, \(x=a\sec { \theta } and\quad y=b\tan { \theta } \)
\(\Rightarrow \quad \frac { x }{ a } =\sec { \theta } \quad and\quad \frac { y }{ b } =\tan { \theta } \quad \quad ..(i)\)
We know that, \(\quad \sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta } \)
\(\therefore \quad { \left( \frac { x }{ a } \right) }^{ 2 }=1+{ \left( \frac { y }{ b } \right) }^{ 2 }\)  [from Eq. (i)]
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) 

LHS = \(\sqrt { \sec ^{ 2 }{ \theta } +{ cosec }^{ 2 }\theta } =\sqrt { \frac { 1 }{ \cos ^{ 2 }{ \theta } } +\frac { 1 }{ \sin ^{ 2 }{ \theta } } } \)
\(=\sqrt { \frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } } } =\sqrt { \frac { 1 }{ \cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } } } \)
\(=\frac { 1 }{ \sin { \theta } \cos { \theta } } \left[ \therefore \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } \left[ \therefore \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { \sin ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } +\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } =\frac { \sin { \theta } }{ \cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \)
\(=\tan { \theta } +\cot { \theta } \)
= RHS

Given, \(\sec ^{ 2 }{ \theta } =x+\frac { 1 }{ 4x } \)
\(\Rightarrow \sec ^{ 2 }{ \theta } ={ x }^{ 2 }+{ \left( \frac { 1 }{ 4x } \right) }^{ 2 }+2.x.\frac { 1 }{ 4x } \)
\(\Rightarrow \sec ^{ 2 }{ \theta } -1={ x }^{ 2 }+\frac { 1 }{ 16{ x }^{ 2 } } +\frac { 1 }{ 2 } -1\)
\(\\ \Rightarrow \sec ^{ 2 }{ \theta } -1={ x }^{ 2 }+\frac { 1 }{ 16{ x }^{ 2 } } +\frac { 1 }{ 2 } -1\Rightarrow \tan ^{ 2 }{ \theta } ={ \left( x-\frac { 1 }{ 4x } \right) }^{ 2 }\)
\(\left[ \because \sec ^{ 2 }{ \theta } -1=\tan ^{ 2 }{ \theta } and\quad { a }^{ 2 }+{ b }^{ 2 }-2ab={ (a-b) }^{ 2 } \right] \)
\(\Rightarrow \tan { \theta } =\pm \left( x-\frac { 1 }{ 4x } \right) \)
\(\Rightarrow \tan { \theta } =x-\frac { 1 }{ 4x } \quad or\quad \tan { \theta } =-\left( x-\frac { 1 }{ 4x } \right) \)
\(\therefore \sec { \theta } +\tan { \theta } =x-\frac { 1 }{ 4x } +x+\frac { 1 }{ 4x } =2x\quad or\quad \sec { \theta } +\tan { \theta } =x+\frac { 1 }{ 4x } -x+\frac { 1 }{ 4x } =2x\)