\(LHS=\frac { \tan { \theta } }{ 1-\cot { \theta } } +\frac { \cot { \theta } }{ 1-\tan { \theta } } \)
\(=\frac { \tan { \theta } }{ 1-\frac { 1 }{ \tan { \theta } } } +\frac { \frac { 1 }{ \tan { \theta } } }{ 1-\tan { \theta } } \left( \because \quad \tan { \theta } =\frac { 1 }{ \cot { \theta } } \right) \)
\(=\frac { \tan ^{ 2 }{ \theta } }{ \tan { \theta } -1 } +\frac { 1 }{ \left( 1-\tan { \theta } \right) \tan { \theta } } \)
\(=\frac { \tan ^{ 3 }{ \theta -1 } }{ \left( \tan { \theta } -1 \right) \tan { \theta } } \left[ \because { a }^{ 3 }-{ b }^{ 3 }=\left( a-b \right) \left( { a }^{ 3 }+{ b }^{ 3 }+ab \right) \right] \)
\(=\frac { \left( \tan { \theta } -1 \right) \left( \tan ^{ 2 }{ \theta } +\tan { \theta } +1 \right) }{ \left( \tan { \theta } -1 \right) \tan { \theta } } =\frac { \tan ^{ 2 }{ \theta } +\tan { \theta } +1 }{ \tan { \theta } } \)
\(=\tan { \theta } +1+\cot { \theta } \)
= RHS

As 3 tan A = 4\(\Rightarrow \tan { A } =\frac { 4 }{ 3 } =\frac { P }{ B } \) 
Using Pythagoras theorem, H² = 4² + 3² \(\Rightarrow\) H = 5
\(\sec { A } =\frac { H }{ B } =\frac { 5 }{ 3 } ,cosecA=\frac { H }{ P } =\frac { 5 }{ 4 } ,\cos { A } =\frac { B }{ H } =\frac { 3 }{ 5 } ,\sin { A } =\frac { P }{ H } =\frac { 4 }{ 5 } \)

Use Pythagoras theorem, to find hypotenuse.
\(\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ a } \)

Let AB=h m be height of tower and BC=30 m be length of its shadow when sun's elevation is 300.Let BD=x m be length of shadow when sun's elevation is 60^o
As ㄥACB=30^o and ㄥADB=60^o
Consider rt . ΔABC, we have
\(\frac { AB }{ BC } \)=tan300
⇒ \(\frac { h }{ 30 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow h=\frac { 30 }{ \sqrt { 3 } } \)
⇒ AB=\(\frac { 30 }{ \sqrt { 3 } } \)m
Consider rt . ΔABD, we have
\(\frac { AB }{ BD } \)=tan600 ⇒ \(\frac { h }{ BD } =\sqrt { 3 } \)
⇒ \(\frac { \frac { 30 }{ \sqrt { 3 } } }{ BD } =\frac { \sqrt { 3 } }{ 1 } \)    [∵ h=\(\frac { 30 }{ \sqrt { 3 } } \)]
∴ Length of shadow=10 m

In ΔAOB and ΔAOC,
AB=AC, AO=AO and OB=OC
∴ ΔAOB≌∆AOC
∴ ㄥBAO=ㄥCAO
⇒ ㄥCAO=1/2ፀ
In right ΔACO,
\(\frac { AO }{ OC } =cosec\left( \frac { \theta }{ 2 } \right) \)
⇒ \(\frac { AO }{ a } =cosec\left( \frac { \theta }{ 2 } \right) \)
⇒ AO=a.cosec\(\left( \frac { \theta }{ 2 } \right) \).sin¢
∴ Height of centre of balloon is a sin¢.cosec\(\left( \frac { \theta }{ 2 } \right) \).