LHS = \(\frac { \cos { A } -\sin { A } +1 }{ \cos { A } +\sin { A } -1 } \)
On dividing numerator and denominator by sin A, we get
\(=\frac { \frac { \cos { A } }{ \sin { A } } -\frac { \sin { A } }{ \sin { A } } +\frac { 1 }{ \sin { A } } }{ \frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } } =\frac { \cot { A } -1+cosecA }{ \cot { A } +1-1cosecA } \) \(\left[ \because \cot { A } =\frac { \cos { \theta } }{ \sin { \theta } } and\frac { 1 }{ \sin { \theta } } =cosec \theta \right] \)
\(=\frac { \cot { A } +cosecA-1 }{ \cot { A } +1-1cosecA } \)
\(=\frac { (\cot { A } +cosecA)-({ cosec }^{ 2 }A-\cot ^{ 2 }{ A } ) }{ \cot { A } +1-1cosecA } \left[ \because 1={ cosec }^{ 2 }A-\cot ^{ 2 }{ A } \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ (\cot { A } +cosecA)(\cot { A } -cosecA) \right] }{ \cot { A } +1-1cosecA } \) \(\left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ 1-(cosecA-\cot { A } ) \right] }{ \cot { A } +1-1cosecA } \)
\(\quad =\frac { (\cot { A } +cosecA)-\left[ 1-cosecA+\cot { A } \right] }{ \cot { A } +1-1cosecA } \)
\(=cosecA+\cot { A } =RHS\)
Hence proved.

In \(\triangle OPQ\),  \(AB\parallel PQ\) [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OB }{ BQ } \)  ... (i)
[by basic proportionality theorem]
Also, in \(\triangle OPR\), \(AC\parallel PR\)  [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OC }{ CR } \)  ... (ii)
From Eqs. (i) and (ii), 
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \Rightarrow \quad BC\parallel QR\)
[by converse of basic proportionality theorem]
Hence proved.

(i) 36  
(ii) 42

(i) Number of logs stacked in each row form a sequence 20, 19, 18, 17,...., which is an AP with first term, a= 20 and common difference, d = 19 - 20 = -1.
Suppose number of rows is n, then S_n= 200
\(\begin{aligned} \Rightarrow \frac{n}{2}[2 \times 20+(n-1)(-1)] & =200 \\ \end{aligned}\)
\(\begin{aligned} {\left[\because S_n\right.} & \left.=\frac{n}{2}\{2 a+(n-1) d\}\right] \end{aligned}\)
\(\begin{array}{lr} \Rightarrow & 400=40 n-n^2+n \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-41 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-25 n-16 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n(n-25)-16(n-25)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & (n-25)(n-16)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n=16 \text { or } n=25 \end{array}\)
Hence, the number of rows is either 25 or 16.
When, n = 16,
a_n= a + (n -1) d = 20 + (16 - 1) ( - 1)
= 20 - 15 = 5
When, n = 25,
a_n = a + (n - 1) d = 20 + (25-1) (-1)
= 20 - 24 = - 4
[\(\because\) number of logs cannot be negative]
Hence, the number of rows is 16 and number of logs in the top row is 5.
(ii) The pattern of logs show space saving creativity, reasoning and balancing.

Let CD be the lighthouse of height h m and let A and B be the two ships on the either side of the lighthouse, such that
∠CBD = 45^o, ∠CAD - 60^o
\(AB=200\left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) m\)
Consider a rt. ΔBDC, ∠D = 90^o, ∠CBD=45^o, we have
\(\frac { CD }{ BD } ={ tan\quad 45 }^{ o }\)
\(\Rightarrow \frac { CD }{ BD } =1\Rightarrow CD=BD\quad ...(i)\)

Again, in rt.ΔADC, ∠D = 90^o, ∠CAD=60^o, we have
\(\frac { CD }{ AD } ={ tan\quad 60 }^{ o }\)
\(\Rightarrow \frac { CD }{ AD } =\sqrt { 3 } \)
\(\Rightarrow CD=\sqrt { 3 } AD\)
\(\Rightarrow AD=\frac { AD }{ \sqrt { 3 } } ...(ii)\)
Adding (i) and (ii), we have
\(AD+BD=\frac { CD }{ \sqrt { 3 } } +CD\)
\(\Rightarrow AB=CD\left( \frac { 1+\sqrt { 3 } }{ \sqrt { 3 } } \right) \)
\(200\left( \frac { 1+\sqrt { 3 } }{ \sqrt { 3 } } \right) =CD\left( \frac { 1+\sqrt { 3 } }{ \sqrt { 3 } } \right) \)
\(\left[ \because AB=\frac { 200\left( \sqrt { 3 } +1 \right) }{ \sqrt { 3 } } \right] \)
⇒ CD=200m
Hence, the height of the lighthouse is 200 m.

Let a be the first term and d be the common difference of the given AP.
Consider, RHS = 3 ( S₂₀ - S₁₀ )
\(=\left[ {20\over 2}\{2a+19d \}-{10\over2}\{2a+9d \}\right]\)
= 3[ 20a + 190d - 10a - 45d ]
= 3[ 10a + 145d ] = 15 [ 2a + 29d ]
\(={30\over2}[2a+(30-1)d]\) = S₃₀ = LHS