Given, \(m\cot { A } =n\)
\(\Rightarrow m.\frac { 1 }{ \tan { A } } =n\quad \left[ \therefore \cot { A } =\frac { 1 }{ \tan { A } } \right] \)
\(\Rightarrow \quad \tan { A } =\frac { m }{ n } \)      ....(i)       
Now, \(\frac { m\sin { A } -n\cos { A } }{ n\cos { A } +m\sin { A } } =\frac { m.\frac { \sin { A } }{ \cos { A } } -n }{ n+m.\frac { \sin { A } }{ \cos { A } } } \)   [dividing numerator and denominator by cosA] 
\(\frac { m\tan { A } -n }{ n+m\tan { A } } =\frac { \frac { { m }^{ 2 } }{ n } -n }{ n+\frac { { m }^{ 2 } }{ n } } \)    [from Eq. (i)]
\(=\frac { \frac { { m }^{ 2 }-{ n }^{ 2 } }{ n } }{ \frac { { m }^{ 2 }+{ n }^{ 2 } }{ n } } =\frac { { m }^{ 2 }-{ n }^{ 2 } }{ { m }^{ 2 }+{ n }^{ 2 } } \) 

\(8\sqrt { 3 } { cosec }^{ 2 }{ 30 }^{ 0 }\sin { { 60 }^{ 0 } } \cos { { 60 }^{ 0 } } { cos }^{ 2 }{ 45 }^{ 0 }\sin { { 45 }^{ 0 } } \tan { { 30 }^{ 0 } } { cosec }^{ 3 }{ 45 }^{ 0 }\)
\(=8\sqrt { 3 } .\frac { 1 }{ \sin ^{ 2 }{ { 30 }^{ 0 } } } .\sin { { 60 }^{ 0 } } .\cos { { 60 }^{ 0 } } .{ cos }^{ 2 }{ 45 }^{ 0 }.\sin { { 45 }^{ 0 } } .\frac { \sin { { 30 }^{ 0 } } }{ \cos { { 30 }^{ 0 } } } .\frac { 1 }{ \sin ^{ 3 }{ { 45 }^{ 0 } } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ \sin ^{ 2 }{ { 30 }^{ 0 } } } .\sin { { 60 }^{ 0 } } .\cos { { 60 }^{ 0 } } .{ cos }^{ 2 }{ 45 }^{ 0 }.\sin { { 45 }^{ 0 } } .\frac { 1 }{ \cos { { 30 }^{ 0 } } } .\frac { 1 }{ \sin ^{ 2 }{ { 45 }^{ 0 } } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ 1/2 } \frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ 2 } { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }.\frac { 1 }{ \sqrt { 3 } /2 } .\frac { 1 }{ { (1/\sqrt { 2 } ) }^{ 2 } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ 4 } .\frac { 2 }{ \sqrt { 3 } } .2=\frac { 24 }{ \sqrt { 3 } } =\frac { 24\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } \)   [on multiplying numerator and denominator by \(\sqrt { 3 } \)]
\(=\frac { 24\times \sqrt { 3 } }{ 3 } =8\sqrt { 3 } \) .

Given,\(\angle A=\angle B={ 45 }^{ 0 }\)
LHS= sin (A + B)
= sin (45⁰ + 45⁰) = sin 90⁰ = 1
and RHS = sin A cos B+cos A sin B
=sin 45⁰ cos 45⁰ + cos 45⁰ sin 45⁰
\(=\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } \)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1\)=LHS
\(\therefore\) sin(A + B) = sin A cos B + cos A sin B
Hence verified.

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30°              (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60°   (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

We have, sin (A + 2B) = \(\frac { \sqrt { 3 } }{ 2 } \)=60⁰   \(\left[ \because \sin { { 60 }^{ 0 } } =\frac { \sqrt { 3 } }{ 2 } \right] \)
\( \Rightarrow\) A + 2B = 60⁰      ....(i)
Again, cos (A + 4B) =0⁰ = cos 90⁰  [cos 90⁰ = 0]
\( \Rightarrow\) A + 4B = 90⁰       ....(ii)
On subtracting Eq. (i) from Eq. (ii), we get
A + 4B = 90⁰
A+2B = 60⁰
2B = 30⁰
\(\Rightarrow \ B=\frac { { 30 }^{ 0 } }{ 2 } ={ 15 }^{ 0 }\)
On substituting B = 15⁰ in Eq. (i), we get
A + 2 x 15⁰=60⁰ \(\Rightarrow\)  A+30⁰ = 60⁰
\( \Rightarrow\) A = 60 ⁰- 30⁰ = 30⁰
Hence, A = 30⁰ and B = 15⁰

Given, \(\angle A=\angle C\therefore \angle A+\angle B+\angle C={ 180 }^{ 0 }\)  [since, sum of three angles of a triangle is equal to 180⁰]
\(\Rightarrow \quad { 90 }^{ 0 }+\angle A+\angle C={ 180 }^{ 0 }\quad \quad [\because \angle B={ 90 }^{ 0 }]\)
\(\Rightarrow \quad \angle A+\angle C={ 90 }^{ 0 }\quad \)
\(\Rightarrow \quad 2\angle A={ 90 }^{ 0 }\)and \(\Rightarrow 2\angle C={ 90 }^{ 0 }\)    \([\therefore \angle A=\angle C]\)
\(\Rightarrow \angle A={ 45 }^{ 0 }\) and \(\Rightarrow \angle C={ 45 }^{ 0 }\)
sin A cos C+cos A sin C
= sin 45⁰ cos 45⁰+cos 45⁰ sin 45⁰
\(=\frac { 1 }{ \sqrt { 2 } } \times \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \times \frac { 1 }{ \sqrt { 2 } } \quad \left[ \because \sin { { 45 }^{ 0 } } =\cos { { 45 }^{ 0 } } =\frac { 1 }{ \sqrt { 2 } } \right] \)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1\) 

Given, sin 3A = cos (A - 26⁰)    ...(i)
where, 3A is an acute angle.
We know that, \(\sin { \theta } =\cos { ({ 90 }^{ 0 }-\theta ) } \)  
From Eq. (i), \(\cos { ({ 90 }^{ 0 }-3A) } =\cos { ({ A-26 }^{ 0 }) } \)
Since, (90⁰ - 3A) and (A - 26⁰) both are acute angles.
90⁰ - 3A = A - 26⁰
\(\Rightarrow\) 4A = 116⁰ \(\Rightarrow A=\frac { { 116 }^{ 0 } }{ 4 } ={ 29 }^{ 0 }\) 

We have, sec 5A = cosec (A + 30⁰)
\(\Rightarrow\) sec 5A = sec[90⁰- (A + 30⁰)]  \([\because sec({ 90 }^{ 0 }-\theta )=cosec\theta ]\)
\(\Rightarrow\) sec 5A = sec(60⁰ - A)
\(\Rightarrow\) 5A = 60⁰ - A  [\(\therefore\) 5A and (60⁰- A) are acute angles]
\(\Rightarrow\) 6A = 60⁰
\(\therefore\) A = 10⁰ 

LHS = \(\sqrt { \frac { cosecA-1 }{ cosecA+1 } } +\sqrt { \frac { cosecA+1 }{ cosecA-1 } } =2\sec { A } \)
\(=\frac { \left[ \sqrt { (cosecA-1) } \sqrt { (cosecA-1) } +\sqrt { (cosecA+1) } \sqrt { (cosecA+1) } \right] }{ \sqrt { (cosecA+1) } \sqrt { (cosecA-1) } } \)
\(=\frac { { \left( \sqrt { cosecA-1 } \right) }^{ 2 }+{ \left( \sqrt { cosecA+1 } \right) }^{ 2 } }{ \sqrt { cosecA+1 } \sqrt { (cosecA-1) } } \)
\(=\frac { \left( cosecA-1 \right) +\left( cosecA-1 \right) }{ \sqrt { { cosec }^{ 2 }A-1 } } \)  \(\left[ \because { \left( \sqrt { a } \right) }^{ 2 }=a\quad and\quad \sqrt { a+b } \times \sqrt { a-b } =\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \right] \)
\(=\frac { 2cosec\quad A }{ \sqrt { \cot ^{ 2 }{ A } -1 } } \left[ \because { cosec }^{ 2 }A=1+\cot ^{ 2 }{ A } \right] \)
\(=\frac { 2cosec\quad A }{ \cot { A } } =\frac { 2 }{ \sin { A } } \times \frac { \sin { A } }{ \cos { A } } \left[ \because cosecA=\frac { 1 }{ \sin { A } } \quad and\cot { A } =\frac { \cos { A } }{ \sin { A } } \right] \)
\(=\frac { 2 }{ \cos { A } } =2\sec { A } \left[ \because \frac { 1 }{ \cos { A } } =\sec { A } \right] \)
= RHS
Hence proved.

LHS = \(\frac { \sin { A } +\cos { A } }{ \sin { A } -\cos { A } } +\frac { \sin { A } -\cos { A } }{ \sin { A } +\cos { A } } \)
\(=\frac { { \left( \sin { A } +\cos { A } \right) }^{ 2 }+{ \left( \sin { A } -\cos { A } \right) }^{ 2 } }{ \left( \sin { A } -\cos { A } \right) \left( \sin { A } +\cos { A } \right) } \)
\(=\frac { \left[ \sin ^{ 2 }{ A } +2\cos { A } \sin { A } +\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } -2\cos { A } \sin { A } +\cos ^{ 2 }{ A } \right] }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \)
\(=\frac { 2\sin ^{ 2 }{ A } +2\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } =\frac { 2(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } ) }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \)
\(=\frac { 2 }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
= RHS
Hence proved.

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

Given, \(a\cos { \theta } -b\sin { \theta } =x\quad ...(i)\)
and \(a\cos { \theta } +b\sin { \theta } =y\quad ....(ii)\)
On squaring Eqs. (i) and (ii) and then adding, we get
\({ x }^{ 2 }+{ y }^{ 2 }={ \left( a\cos { \theta } -b\sin { \theta } \right) }^{ 2 }+{ \left( a\cos { \theta } +b\sin { \theta } \right) }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } -2ab\cos { \theta } \sin { \theta } +{ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } +2ab\cos { \theta } \sin { \theta } \)\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab\quad and\quad { (a-b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }-2ab\quad \right] \)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } +{ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } \)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }(\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } )+{ b }^{ 2 }(\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } )\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right] \)
Hence proved.

Given, \(\tan { \theta } +\sin { \theta } =m\quad \quad \quad ...(i)\)
and \(\tan { \theta } -\sin { \theta } =n\quad \quad ...(ii)\)
On adding Eqs. (i) and (ii), we get
\(2\tan { \theta } =m+n\quad \Rightarrow \tan { \theta } =\frac { m+n }{ 2 } \)
\(\therefore \quad \cot { \theta } =\frac { 1 }{ \tan { \theta } } =\frac { 2 }{ m+n } \quad \quad ...(iii)\)
On subtracting Eq. (ii) from Eq. (i), we get, \(2\sin { \theta } =m-n\)
\(\Rightarrow \sin { \theta } =\frac { m-n }{ 2 } \quad \quad \quad \quad .....(iv)\)
\(\therefore \quad cosec\theta =\frac { 1 }{ \sin { \theta } } =\frac { 2 }{ m-n } \)
We know that, \({ cosec }^{ 2 }\theta -\cot ^{ 2 }{ \theta } =1\)
\({ \left( \frac { 2 }{ m-n } \right) }^{ 2 }-{ \left( \frac { 2 }{ m+n } \right) }^{ 2 }=1\)   [from Eqs. (ii) and (iv)]
\(\Rightarrow \frac { 4 }{ { \left( m-n \right) }^{ 2 } } -\frac { 4 }{ { \left( m+n \right) }^{ 2 } } =1\)
\(\Rightarrow 4\left[ \frac { 1 }{ { \left( m-n \right) }^{ 2 } } -\frac { 1 }{ { \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { { \left( m+n \right) }^{ 2 }-{ \left( m-n \right) }^{ 2 } }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { ({ m }^{ 2 }+{ n }^{ 2 }+2mn)-({ m }^{ 2 }+{ n }^{ 2 }-2mn) }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { 2mn+2mn) }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow \frac { 16mn }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } =1\)
\(\Rightarrow \frac { 16mn }{ { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }^{ 2 } } =1\)
\(\Rightarrow { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }^{ 2 }=16mn\)
\(\therefore \quad { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }=4\sqrt { mn } \)   [taking positive square roor]
Hence proved. 

LHS = \(\cos { \theta } \sin { \theta } -\frac { \sin { \theta } \sin { \theta } \cos { \theta } }{ cosec\theta } -\frac { \cos { \theta } \cos { \theta } \sin { \theta } }{ \sec { \theta } } +\sec { \theta } \)
\(=\cos { \theta } \sin { \theta } -\sin ^{ 3 }{ \theta } \cos { \theta } -\cos ^{ 3 }{ \theta } \sin { \theta } +\sec { \theta } \)
\(\\ =\cos { \theta } \sin { \theta } -\sin { \theta } \cos { \theta } (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } )+\sec { \theta } \)

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

LHS = \(\sqrt { \frac { 1-\cos { \theta } }{ 1+\cos { \theta } } } =\sqrt { \frac { 1-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } } } \left( \because \quad \cos { { 60 }^{ ° } } =\frac { 1 }{ 2 } \right) \)
\(=\sqrt { \frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
RHS \(=\frac { \sin { \theta } }{ 1+\cos { \theta } } =\frac { \sin { { 60 }^{ ° } } }{ 1+\cos { { 60 }^{ ° } } } \)
\(=\frac { \frac { \sqrt { 3 } }{ 2 } }{ 1+\frac { 1 }{ 2 } } =\frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 3 }{ 2 } } \)
\(=\frac { 1 }{ \sqrt { 3 } } =LHS\)
Hence relation is verified for \(\theta ={ 60 }^{ ° }\)

\(\cos { \theta } +\sin { \theta } =\sqrt { 2 } \cos { \theta } \)
\(\Rightarrow \quad \sin { \theta } =\cos { \theta } \left( \sqrt { 2 } -1 \right) \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( \sqrt { 2 } -1 \right) \left( \sqrt { 2 } +1 \right) }{ \left( \sqrt { 2 } +1 \right) } \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( 2-1 \right) }{ \sqrt { 2 } +1 } \)
\(\Rightarrow \quad \left( \sqrt { 2 } +1 \right) \sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \sqrt { 2 } \sin { \theta } +\sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta } \)

\(LHS=\frac { \cos { A } }{ 1-\tan { A } } +\frac { \sin { A } }{ 1-\cot { A } } \)
\(=\frac { \cos { A } }{ 1-\left( \frac { \sin { A } }{ \cos { A } } \right) } +\frac { \sin { A } }{ 1-\left( \frac { \cos { A } }{ \sin { A } } \right) } \)
\(=\frac { \cos ^{ 2 }{ A } }{ \cos { A } .\sin { A } } +\frac { \sin ^{ 2 }{ A } }{ \sin { A } .\cos { A } } \)
\(=\frac { \cos ^{ 2 }{ A } }{ \cos { A } -\sin { A } } .\frac { \sin ^{ 2 }{ A } }{ \cos { A } -\sin { A } } \)
\(=\frac { \cos ^{ 2 }{ A.\sin ^{ 2 }{ A } } }{ \cos { A } .\sin { A } } \)
\(=\frac { \left( \cos { A } -\sin { A } \right) \left( \cos { A } +\sin { A } \right) }{ \left( \cos { A } -\sin { A } \right) } \)
\(=\cos { A } +\sin { A } \)
= RHS

\(b\cos { \theta } =a\)
\(\Rightarrow \quad \cos { \theta } =\frac { a }{ b } \)
\(cosec\theta =\frac { b }{ \sqrt { { b }^{ 2 }-{ a }^{ 2 } } } ,\cot { \theta } =\frac { a }{ \sqrt { { b }^{ 2 }-{ a }^{ 2 } } } \)
\(cosec\theta +\cot { \theta } =\frac { b+a }{ \sqrt { { b }^{ 2 }-{ a }^{ 2 } } } =\sqrt { \frac { b+a }{ b-a } } \)

\(\frac { { cosec }^{ 2 }\theta }{ { cosec }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { cosec }\theta +1 } \)
\(={ cosec }^{ 2 }\theta \left[ \frac { 1 }{ \frac { 1 }{ \sin { \theta } } -1 } -\frac { 1 }{ \frac { 1 }{ \sin { \theta } } +1 } \right] \)
\(={ cosec }^{ 2 }\theta \left[ \frac { \sin { \theta } }{ 1-\sin { \theta } } -\frac { \sin { \theta } }{ 1+\sin { \theta } } \right] \)
\(=\frac { 1\times \sin { \theta } }{ \sin ^{ 2 }{ \theta } } \left[ \frac { \left( 1+\sin { \theta } \right) -\left( 1-\sin { \theta } \right) }{ \left( 1-\sin { \theta } \right) \left( 1+\sin { \theta } \right) } \right] \)
\(=\frac { 1 }{ \sin { \theta } } \left[ \frac { 2\sin { \theta } }{ 1-\sin ^{ 2 }{ \theta } } \right] \)
\(=\frac { 2 }{ \cos ^{ 2 }{ \theta } } =2\sec ^{ 2 }{ \theta } \)
= RHS

\(LHS=\frac { 1 }{ cosecA-\cot { A } } -\frac { 1 }{ \sin { A } } \)
\(=\frac { 1 }{ cosecA-\cot { A } } \times \left( \frac { cosecA+\cot { A } }{ cosecA+\cot { A } } \right) -cosecA\)
\(\left( \because \quad \sin { A } =\frac { 1 }{ cosecA } \right) \)
\(=\frac { cosecA+\cot { A } }{ { cosec }^{ 2 }A-\cot ^{ 2 }{ A } } -cosecA\)
\(=\frac { cosecA+\cot { A } }{ 1 } -cosecA=\cot { A } \)
Now RHS = \(\frac { 1 }{ \sin { A } } -\frac { 1 }{ cosecA+\cot { A } } \)
\(=cosecA-\frac { 1 }{ cosecA+\cot { A } } \times \left( \frac { cosecA-\cot { A } }{ cosecA-\cot { A } } \right) \)
\(=cosecA-\frac { \left( cosecA-\cot { A } \right) }{ { cosec }^{ 2 }A-\cot ^{ 2 }{ A } } \)
\(=cosecA-\frac { \left( cosecA-\cot { A } \right) }{ 1 } \)
= cot A
\(\therefore \quad LHS=RHS\)

Let \(\sec { \theta } +\tan { \theta } =\lambda \)      ....(i)
We know that  \(\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } =1\)
\(\Rightarrow \quad \left( \sec { \theta } +\tan { \theta } \right) \left( \sec { \theta } -\tan { \theta } \right) =1\)
\(\Rightarrow \quad \lambda \left( \sec { \theta } -\tan { \theta } \right) =1\)
\(\Rightarrow \quad \sec { \theta } -\tan { \theta } =\frac { 1 }{ \lambda } \)          .... (ii)
Adding eqns.(i) and (ii),
\(2\sec { \theta } =\lambda +\frac { 1 }{ \lambda } \)
\(\Rightarrow \quad 2\left( x+\frac { 1 }{ 4x } \right) =\lambda +\frac { 1 }{ \lambda } \)
\(\Rightarrow \quad 2x+\frac { 1 }{ 2x } =\lambda +\frac { 1 }{ \lambda } \)
Comparing both sides,
\(\lambda =2x\quad or\quad \lambda =\frac { 1 }{ 2x } \)
\(\Rightarrow \quad \sec { \theta } +\tan { \theta } =2x\quad or\quad \frac { 1 }{ 2x } \)

sin² A + cos² A = 1
(i) \(\sin { A } =\sqrt { 1-\cos ^{ 2 }{ A } } \)
\(=\sqrt { 1-\frac { 1 }{ \sec ^{ 2 }{ A } } } \)
\(=\sqrt { \frac { \sec ^{ 2 }{ A } -1 }{ \sec ^{ 2 }{ A } } } =\sqrt { \frac { \sec ^{ 2 }{ A } -1 }{ \sec { A } } } \)
(ii) \(\tan { A } =\frac { \sin { A } }{ \cos { A } } =\sin { A } \sec { A } \)
\(=\sqrt { \frac { \sec ^{ 2 }{ A } -1 }{ \sec ^{ 2 }{ A } } } \times \sec { A } =\sqrt { \sec ^{ 2 }{ A } -1 } \)
(iii) \(cosecA=\frac { 1 }{ \sin { A } } =\frac { \sec { A } }{ \sqrt { \sec ^{ 2 }{ A } -1 } } \)

In \(\triangle\) OPQ, we have
OQ² = OP² + PQ²
i.e., (1 + PQ)² = OP² + PQ²
i.e., 1 + PQ² + 2PQ = OP² + PQ²
i.e., 1 + 2PQ = 7²
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
So,    \(\sin \mathrm{Q}=\frac{7}{25} \text { and } \cos \mathrm{Q}=\frac{24}{25}\)

2tan²45° + cos²30° − sin²60°
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2+\frac{3}{4}-\frac{3}{4}=2\) 

\(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
\(=\frac{\frac{1}{2}+\frac{1}{1}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+\frac{1}{1}}=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}\)
\(\begin{aligned} & {\left[\begin{array}{c} \because \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \operatorname{cosec} 60^{\circ}=\sec 30^{\circ}=\frac{2}{\sqrt{3}}, \\ \cot 45^{\circ}=\tan 45^{\circ}=1 \end{array}\right]} \\ \end{aligned}\)
\(=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}} \times \frac{4-3 \sqrt{3}}{4-3 \sqrt{3}}\)
[multiplying numerator and denominator by the conjugate of 4 + 3\(\sqrt3\), i.e. 4 - 3\(\sqrt3\)]
\(\begin{aligned} & =\frac{12 \sqrt{3}-27-16+12 \sqrt{3}}{(4)^2-(3 \sqrt{3})^2}\left[\because(a+b)(a-b)=a^2-b^2\right] \\ \end{aligned}\)
\(\begin{aligned} =\frac{24 \sqrt{3}-43}{16-27}=\frac{-(43-24 \sqrt{3})}{-11}=\frac{43-24 \sqrt{3}}{11} \end{aligned}\)

LHS = sec A (1 – sin A)(sec A + tan A) \(=\left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\)
\(=\frac{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}{\cos ^{2} \mathrm{~A}}=\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}\)
\(=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=1=\mathrm{RHS}\) 

Given, cos A = \(\frac{5}{13}=\frac{\operatorname{Base}(B)}{\text { Hypotenuse }(H)}\)
By Pythagoras theorem, we get
H² = P² + B²
\(\Rightarrow\) (13)² = P² + (5)²
\(\Rightarrow\) 169 = P² + 25
\(\Rightarrow\)  169 - 25 = P²
\(\Rightarrow\)   144 = P²
\(\Rightarrow\)     P = 12
Now, we have to verify that
\(\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\cos A+\sin A\)
\(\begin{aligned} \text { LHS } & =\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A} \end{aligned}\)
\(\begin{aligned} =\frac{\frac{5}{13}}{1-\frac{12}{5}}+\frac{\frac{12}{13}}{1-\frac{5}{12}} \end{aligned}\)
\(=\frac{\frac{5}{13}}{\frac{5-12}{5}}+\frac{\frac{12}{13}}{\frac{12-5}{12}}=\frac{5}{13} \times\left(-\frac{5}{7}\right)+\frac{12}{13} \times \frac{12}{7}\)
\(=-\frac{25}{91}+\frac{144}{91}=\frac{119}{91}=\frac{17}{13}\)
RHS = cosA + sinA
\(=\frac{5}{13}+\frac{12}{13}=\frac{17}{13}\)    
LHS = RHS            Hence proved.

\(\begin{aligned} \mathrm{LHS} & =\frac{\sin A+\cos A}{\sin A-\cos A}+\frac{\sin A-\cos A}{\sin A+\cos A} \\ \end{aligned}\)
\(\begin{aligned} =\frac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)} \end{aligned}\)
\(=\frac{\left[\begin{array}{c} \sin ^2 A+2 \sin A \cos A+\cos ^2 A+\sin ^2 A \\ -2 \sin A \cos A+\cos ^2 A \end{array}\right]}{\sin ^2 A-\cos ^2 A}\)
\(\left[\because(a \pm b)^2=a^2+b^2 \pm 2 a b\right]\)
\(\begin{aligned} & =\frac{2 \sin ^2 A+2 \cos ^2 A}{\sin ^2 A-\cos ^2 A}=\frac{2\left(\sin ^2 A+\cos ^2 A\right)}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\cos ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-\left(1-\sin ^2 A\right)} \end{aligned}\)
\(\begin{aligned} =\frac{2}{\sin ^2 A-1+\sin ^2 A} \end{aligned}\)
\(\begin{aligned} =\frac{2}{2 \sin ^2 A-1} \end{aligned}\)
= RHS                                   Hence proved.

To prove (cosec \(\theta\) - sin \(\theta\))(sec \(\theta\) - cos \(\theta\)) (tan \(\theta\) + cot \(\theta\)) = 1
Proof LHS = (cosec \(\theta\) - sin \(\theta\)) (sec \(\theta\) - cos \(\theta\)) (tan \(\theta\) + cot \(\theta\))
\(=\left(\frac{1}{\sin \theta}-\sin \theta\right)\left(\frac{1}{\cos \theta}-\cos \theta\right)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
\(\left[\begin{array}{l} \because \operatorname{cosec} A=\frac{1}{\sin A}, \sec A=\frac{1}{\cos A} \\ \tan A=\frac{\sin A}{\cos A}, \cot A=\frac{\cos A}{\sin A} \end{array}\right]\)
\(=\frac{1-\sin ^2 \theta}{\sin \theta} \times \frac{1-\cos ^2 \theta}{\cos \theta} \times \frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\)
[ \(\because\) sin² A + cos² A = 1 \(\Rightarrow\) 1 - cos² A = sin² A and 1 - sin² A = cos² A]
\(=\frac{\cos ^2 \theta}{\sin \theta} \times \frac{\sin ^2 \theta}{\cos \theta} \times \frac{1}{\cos \theta \sin \theta}=1=\text { RHS }\)
                                         Hence proved.