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

In \(\Delta AOP\), tan 30°=\(\frac{H-120}{OP}\)
\(\frac { 1 }{ \sqrt { 3 } } =\)\(\frac{H-120}{OP}\) 

OP = (H -120)\(\sqrt{3}\) ...(i)
In \(\Delta OPA'\), tan 60°=\(\frac{H+120}{OP}\)
OP =\(\frac { H+120 }{ \sqrt { 3 } } \)...(ii)
From (i) and (ii), we get
\(\frac { H+120 }{ \sqrt { 3 } } \)=\(\sqrt{3}\)(H-120)
So height of cloud H = 240 m.

\(\frac { 1 }{ p } =\frac { 1 }{ \sec { \theta } +\tan { \theta } } \times \frac { \left( \sec { \theta } -\tan { \theta } \right) }{ \sec { \theta } -\tan { \theta } } \)
\(\frac { 1 }{ p } =\frac { \sec { \theta } -\tan { \theta } }{ \sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } } \times \sec { \theta } -\tan { \theta } \)
Solving \(\sec { \theta } +\tan { \theta } =p,\) and \(\sec { \theta } -\tan { \theta } =\frac { 1 }{ p } \)
we get  \(\sec { \theta } =\frac { 1 }{ 2 } \left( p+\frac { 1 }{ p } \right) =\frac { { p }^{ 2 }+1 }{ 2p } \)
and \(\tan { \theta } =\frac { 1 }{ 2 } \left( p-\frac { 1 }{ p } \right) =\frac { { p }^{ 2 }-1 }{ 2p } \)
\(\therefore \quad \cos { \theta } =\frac { 2p }{ { p }^{ 2 }+1 } \) and \(\sin { \theta } =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Alternative Method :
Adding eq.(i) and eq.(ii), 2
\(\sec { \theta } +\tan { \theta } =p,\)   ...(i)
\(\sec { \theta } -\tan { \theta } =\frac { 1 }{ p } \) ..(ii)
\(2\sec { \theta } =p+\frac { 1 }{ p } \)
\(\Rightarrow \quad 2\sec { \theta } =\frac { { p }^{ 2 }+1 }{ p } \)
\(\Rightarrow \quad \sec { \theta } =\frac { { p }^{ 2 }+1 }{ 2p } \)
\(\Rightarrow \quad \frac { 1 }{ \cos { \theta } } =\frac { { p }^{ 2 }+1 }{ 2p } \)
\(\Rightarrow \quad \cos { \theta } =\frac { 2p }{ { p }^{ 2 }+1 } \)
On subtracting eq.(ii) from eq.(i)
\(\sec { \theta } +\tan { \theta } =p,\)
\(\sec { \theta } -\tan { \theta } =\frac { 1 }{ p } \)
\( 2\tan { \theta } =p-\frac { 1 }{ p } \)
\(\tan { \theta } =\frac { { p }^{ 2 }-1 }{ 2p } \)
\(\tan { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \)
\(\Rightarrow \quad \sin { \theta } =\tan { \theta } .\cos { \theta } \)
\(=\frac { { p }^{ 2 }-1 }{ 2p } \times \frac { 2p }{ { p }^{ 2 }+1 } \)
\(=\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

\(\tan { \left( A+B \right) } =\frac { \tan { A } +\tan { B } }{ 1-\tan { A } \tan { B } } \)
(i) tan 75° = tan (45° + 30°)
\(=\frac { \tan { { 45 }^{ ° } } +\tan { { 30 }^{ ° } } }{ 1.\tan { { 45 }^{ ° } } .\tan { { 30 }^{ ° } } } \)
\(=\frac { 1+\frac { 1 }{ \sqrt { 3 } } }{ 1-\frac { 1 }{ \sqrt { 3 } } } \)
\(\therefore \quad \tan { { 75 }^{ ° } } =\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \)
(ii) tan 90° = tan (60° + 30°)
\(=\frac { \tan { { 60 }^{ ° } } +\tan { { 30 }^{ ° } } }{ 1-\tan { { 60 }^{ ° } } \tan { { 30 }^{ ° } } } \)
\(=\frac { \sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } }{ 1-\sqrt { 3 } \times \frac { 1 }{ \sqrt { 3 } } } \)
\(=\frac { \frac { 3+1 }{ \sqrt { 3 } } }{ 0 } \)
\(\therefore \quad \tan { { 90 }^{ ° } } =\infty \)

In right angled \(\triangle PQR,\)
PQ = 12 cm, PR = 13 cm   [given]
Then, PQ² + QR² = PR²   [by using Pythagoras theorem]
\(\Rightarrow\)  (12)² + QR² = (13)² 
\(\Rightarrow\)  144 + QR² = 169
\(\Rightarrow\)  QR² = 169 - 144 = 25
\(\Rightarrow\)  QR = 5 [taking positive square root since, side cannot be negative]
Now, \(tanP=\frac { P }{ B } =\frac { QR }{ PQ } =\frac { 5 }{ 12 } \)
\(and\quad cot \quad R=\frac { B }{ P } =\frac { QR }{ PQ } =\frac { 5 }{ 12 } \)
\(\therefore \quad tanP-cosR=\frac { 5 }{ 12 } - \frac { 5 }{ 12 } =0\) 

\(\angle QPR={ 60 }^{ 0 }\quad and\quad \angle PRQ={ 30 }^{ 0 }.\)

Given, \(\tan { A } =a\tan { B } \Rightarrow \tan { B } =\frac { 1 }{ a } \tan { A } \)
\(\Rightarrow \cot { A } =\frac { a }{ \tan { A } } \quad \quad ..(i)\)
and \(\sin { A } =b\sin { B } \Rightarrow \sin { B } =\frac { 1 }{ b } \sin { A } \)
\(\Rightarrow cosecB=\frac { b }{ \sin { A } } \quad \left[ \because cosec\theta =\frac { 1 }{ \sin { \theta } } \right] \quad ...(ii)\)
We know that, \({ cosec }^{ 2 }B-\cot ^{ 2 }{ B } =1\)
\(\Rightarrow \frac { { b }^{ 2 } }{ \sin ^{ 2 }{ A } } -\frac { { a }^{ 2 } }{ \tan ^{ 2 }{ A } } =1\)  [from Eqs. (i) and (ii)]
\(\Rightarrow \frac { { b }^{ 2 } }{ \sin ^{ 2 }{ A } } -\frac { { a }^{ 2 }\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } \quad \left[ \because \tan { A } =\frac { \sin { A } }{ \cos { A } } \right] \)
\(\Rightarrow \frac { { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } } =1\)
\(\Rightarrow { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } =\sin ^{ 2 }{ A } \)
\(\Rightarrow { b }^{ 2 }-{ a }^{ 2 }\cos ^{ 2 }{ A } =1-\cos ^{ 2 }{ A } \)
\(\Rightarrow \quad { b }^{ 2 }-1={ a }^{ 2 }\cos ^{ 2 }{ A } -\cos ^{ 2 }{ A } \)
\(\Rightarrow \frac { { b }^{ 2 }-1 }{ { a }^{ 2 }-1 } =\cos ^{ 2 }{ A } \)
Hence proved.

LHS = \(\quad \sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } } =\sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } \times \frac { 1+\sin { A } }{ 1+\sin { A } } } \)  [on multiplying numerator and denominator by \(\sqrt { (1+\sin { A } ) } \)]
\(=\sqrt { \frac { { (1+\sin { A } ) }^{ 2 } }{ 1-\sin ^{ 2 }{ A } } } =\sqrt { \frac { { (1+\sin { A } ) }^{ 2 } }{ \cos ^{ 2 }{ A } } } \)  \(\left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1\Rightarrow 1-\sin ^{ 2 }{ A } =\cos ^{ 2 }{ A } \right] \)
\(=\frac { 1+\sin { A } }{ \cos { A } } =\frac { \sin { A } }{ \cos { A } } +\frac { 1 }{ \cos { A } } \)
\(=\sec { A } +\tan { A } \quad \left[ \because \sec { A } =\frac { 1 }{ \cos { A } } ,\tan { A } =\frac { \sin { A } }{ \cos { A } } \right] \)
= RHS
Hence proved.

LHS = \(\tan { { 48 }^{ 0 } } \tan { { 23 }^{ 0 } } \tan { { 42 }^{ 0 } } \tan { { 67 }^{ 0 } }\)
\(=\tan { { 48 }^{ 0 } } \tan { { 23 }^{ 0 } } \tan { ({ 90 }^{ 0 }-{ 48 }^{ 0 }) } \tan { ({ 90 }^{ 0 }-{ 23 }^{ 0 }) } \)
\(=\tan { { 48 }^{ 0 } } \tan { { 23 }^{ 0 } } \cot { { 48 }^{ 0 } } \cot { { 23 }^{ 0 } } \quad \left[ \because \tan { ({ 90 }^{ 0 }-\theta ) } =\cot { \theta } \right] \)
\(=\tan { { 48 }^{ 0 } } \tan { { 23 }^{ 0 } } \frac { 1 }{ \tan { { 48 }^{ 0 } } } \frac { 1 }{ \tan { { 23 }^{ 0 } } } \)
= 1 = RHS

First term
\(=\sin { A } \left( \frac { 1-\cos { A } +1+\cos { A } }{ 1-\sin ^{ 2 }{ A } } \right) =\sin { A } .\frac { 2 }{ \sin ^{ 2 }{ A } } \)
\(=\frac { 2 }{ \sin { A } } =cosecA\quad \quad \left[ \therefore \frac { 1 }{ \sin { A } } =cosecA \right] \)
= Third term
Second term =\(\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } } +\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } } \)
\(=\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } \times \frac { 1+\cos { A } }{ 1+\cos { A } } } +\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } \times \frac { 1-\cos { A } }{ 1-\cos { A } } } \)
\(=\frac { 1+\cos { A } }{ \sqrt { \sin ^{ 2 }{ A } } } +\frac { 1-\cos { A } }{ \sqrt { \sin ^{ 2 }{ A } } } \quad [\because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1]\)
\(=\frac { 1+\cos { A } +1-\cos { A } }{ \sin { A } } \)
= Third term  \([\because \frac { 1 }{ \sin { A } } =cosecA]\)
Hence proved. 

LHS =  \(\frac { \frac { 1 }{ \cos ^{ 2 }{ \theta } } -\sin ^{ 2 }{ \theta } }{ \frac { \sin ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } } } =\frac { 1-\sin ^{ 2 }{ \theta } .\cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } } ={ cosec }^{ 2 }\theta -\cos ^{ 2 }{ \theta } \)
\(=1+\cot ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } \) 

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As, \(\sqrt { 3 } \sin { \theta } =\cos { \theta } \Rightarrow \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \frac { \sin { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \sin { \theta } +\cos { \theta } } =\frac { \tan { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \tan { \theta } +1 } \)
\(\frac { 1+\sqrt { 3 } }{ 4 } \)

Use Pythagoras theorem, to find base
\(\frac {5}{13}\)