\( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} =\sec \theta+\tan \theta \)
\(\mathbf{L H S} =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}\)
[Multiplying the Numerator and denominator by \(\sqrt{1-\sin \theta}\)]
\( =\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\sqrt{\frac{\cos ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{\cos \theta}{1-\sin \theta} \)
\(=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
[Multiplying Numerator and denominator by \(1+\sin \theta\)]
\( =\frac{\cos \theta(1+\sin \theta)}{1^{2}-\sin ^{2} \theta}=\frac{\cos \theta(1+\sin \theta)}{\cos ^{2} \theta} \)
\({\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \)
\(=\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta+\tan \theta=\text { RHS }\)

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

\(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } = \frac { \frac { 1 }{ cos\theta } }{ sin\theta } -\frac { sin\theta }{ cos\theta } =\frac { 1 }{ sin\theta cos\theta } -\frac { sin\theta }{ cos\theta } \)
\(=\frac { 1-si{ n }^{ 2 }\theta }{ sin\theta cos\theta } =cot\theta \)

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

sec\(\theta \) - cos\(\theta \)  = \(\frac { 1 }{ cos\theta } -cos\theta =\frac { 1-co{ s }^{ 2 }\theta }{ cos\theta } \)
= \(\frac { si{ n }^{ 2 }\theta }{ cos\theta } \) [since 1 - cos²\(\theta \) = sin²\(\theta \)]
= \(\frac { sin\theta }{ cos\theta } \times sin\theta =tan\theta sin\theta \) 

1 + \(\frac { co{ t }^{ 2 }\theta }{ 1+cosec\theta +1 } \) = 1+ \(\frac { cose{ c }^{ 2 }\theta -1 }{ cosec\theta +1 } \) [since cosec²-1 = cot²\(\theta \)
= 1+\(\frac { (cosec\theta +1)(cosec\theta -1) }{ cosec\theta +1 } \) 
1 +( cosec\(\theta \)-1) = cosec\(\theta \)

\(\frac { sinA }{ 1+cosA } = \)\(\frac { sinA }{ 1+cosA } \)\(\times \frac { 1-cosA }{ 1-cosA } \) [ multiply numerator and denominator by the conjugate of 1+cosA]
= \(\frac { sinA(1-cosA) }{ (1+cosA)\quad (1-cosA) } =\frac { sinA(1-cosA) }{ 1-co{ s }^{ 2 }A } \) 
= \(\frac { sinA(1-cosA) }{ si{ n }^{ 2 }A } =\frac { 1-cosA }{ sinA } \) 

tan² \(\theta \) - sin ²\(\theta \) = tan² \(\theta \) -\(\frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \),cos²\(\theta \) 
= tan² \(\theta \) (1-cos² \(\theta \) ) = tan² \(\theta \) sin^{2 \(\theta \)}

Given \(\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a} \)
\(\therefore a=\frac{1+\sin \theta}{\cos \theta} \)
\(\mathrm{LHS}=\frac{a^{2}-1}{a^{2}+1} \)
\(=\frac{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}-1}{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}+1}\)
\(=\frac{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}-1}{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}+1}\)
\(=\frac{\frac{1+\sin ^{2} \theta+2 \sin \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{\cos ^{2} \theta}}\)
\(=\frac{\left(1-\cos ^{2} \theta\right)+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{1+\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta}\)
\(=\frac{\sin ^{2} \theta+\sin ^{2} \theta+2 \sin \theta}{1+1+2 \sin \theta}\)
\(=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2+2 \sin \theta} \)
\(=\frac{2 \sin \theta(\sin \theta+1)}{2(1+\sin \theta)} \)
= sin \(\theta \) = RHS

sin \(\theta \) + cos \(\theta \) = P and sec \(\theta \) + cosec \(\theta \) = 9
LHS = q(p² - 1)
= (sec \(\theta \) + cosec \(\theta \)) [(sin \(\theta \) + cos \(\theta \))² -1]
\(=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1\right) \)
\(=\left(\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\right)(1+2 \sin \theta \cos \theta-1) \)
\(=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta} \times 2 \sin \theta \cos \theta \)
\(=2(\sin \theta+\cos \theta)=2 p=\text { RHS } \)

Given
\( \frac{\cos \alpha}{\cos \beta}=m \)
\(\frac{\cos \alpha}{\sin \beta} =n \)
\(\text { LHS } =\left(m^{2}+n^{2}\right) \cos ^{2} \beta \)
\(=\left(\frac{\cos ^{2} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}\right) \cos ^{2} \beta \)
\(=\frac{\left(\cos ^{2} \alpha \sin ^{2} \beta+\cos ^{2} \alpha \cos ^{2} \beta\right)}{\cos ^{2} \beta \sin ^{2} \beta} \cos ^{2} \beta \)
\(=\frac{\cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)}{\sin ^{2} \beta} \)
\(=\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}(1) \)
\(=\left(\frac{\cos \alpha}{\sin \beta}\right)^{2} \)
= n² = RHS

We have sin \(\theta \) + cos \(\theta \) = \(\sqrt{3}\)
Squaring on both the sides,
\((\sin \theta+\cos \theta)^{2} =(\sqrt{3}) \)
\(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta =3 \)
1 + 2 sin \(\theta \) cos \(\theta \) = 3
2 sin \(\theta \) cos \(\theta \) = 3 - 1
2 sin \(\theta \) cos \(\theta \) = 2
sin \(\theta \) cos \(\theta \) = 2/2
sin \(\theta \) cos \(\theta \) = 1
Now to prove tan \(\theta \) + cot \(\theta \) = 1
\( \mathrm{LHS} =\tan \theta+\cot \theta \)
\(=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta} \)
\(=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta} \)
\(=\frac{1}{1}\)
= 1
tan \(\theta \) + cot \(\theta \)  = 1

We have \(\frac { co{ s }^{ 2 }\theta }{ sin\theta } \) =p           ...(1) and \(\frac { sin^{ 2 }\theta }{ cos\theta } \) =q.         ..(2)
p²q²(p² + q² + 3)= \({ \left( \frac { co{ s }^{ 2 }\theta }{ sin\theta } \right) }^{ 2 }{ \left( \frac { sin^{ 2 }\theta }{ cos\theta } \right) }^{ 2 }\times \left[ { \left( \frac { co{ s }^{ 2 }\theta }{ sin\theta } \right) }^{ 2 }+{ \left( \frac { si{ n }^{ 2 }\theta }{ cos\theta } \right) }^{ 2 }+3 \right] \) [from (1) and (2)]
=\(\left( \frac { co{ s }^{ 4 }\theta }{ si{ n }^{ 2 }\theta } \right) \left( \frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \right) \times \left[ \frac { co{ s }^{ 4 }\theta }{ si{ n }^{ 2 }\theta } +\frac { si{ n }^{ 4 }\theta }{ co{ s }^{ 2 }\theta } +3 \right] \)
= (cos²\(\theta \)\(\times \)sin²\(\theta \))\(\times \) \(\left[ \left( \frac { co{ s }^{ 6 }\theta +si{ n }^{ 6 }\theta +3si{ n }^{ 2 }\theta co{ s }^{ 2 }\theta }{ si{ n }^{ 2 }\theta co{ s }^{ 2 }\theta } \right) \right] \)
= cos⁶\(\theta \) + sin⁶\(\theta \) + 3sin²\(\theta \)cos²\(\theta \)
= (cos²\(\theta \))³ + (sin²\(\theta \))³ + 3sin²\(\theta \)cos²\(\theta \)
= [(cos²\(\theta \) + sin²\(\theta \))³ - 3cos²\(\theta \) sin²\(\theta \)(cos²\(\theta \)  + sin²\(\theta \))] + 3sin²\(\theta \)cos²\(\theta \)
= 1 - 3cos²\(\theta \)sin²\(\theta \)(1) + 3cos²\(\theta \)sin²\(\theta \)  = 1

Given cosec\(\theta \) + cot\(\theta \) = p   ...(1)
cosec²\(\theta \) - cot²\(\theta \) = 1 (identity)
\(\operatorname{cosec} \theta-\cot \theta=\frac{1}{\operatorname{cosec} \theta+\cot \theta}\)
cosec\(\theta \) - cot\(\theta \) =\(\frac { 1 }{ { p } } \)  .... (2)
Adding(1) and (2) we get, 2cosec\(\theta \) = \(p+\frac { 1 }{ p } \) 
2cosec\(\theta \)\(\frac { { p }^{ 2 }+1 }{ p } \) ....(3) 
Subtracting (2) from (1), we get, 2cot\(\theta \) = \(p-\frac { 1 }{ p } \) 
2cot\(\theta \) = \(\frac { { p }^{ 2 }-1 }{ p } \)   ...(4)
Dividing (4) by (3) we get,\(\frac { 2cot\theta }{ 2cosec\theta } =\frac { { p }^{ 2 }-1 }{ p } \times \frac { p }{ { p }^{ 2 }+1 } gives,cos\theta =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

\(\frac { sinA }{ 1+cosA } +\frac { sinA }{ 1-cosA } \)
\(=\frac { sinA(1-cosA)+sinA(1+cosA) }{ (1+cosA)(1-cosA) } \)
\(=\frac { sinA-sinAcosA+sinA+sinAcosA }{ 1-co{ s }^{ 2 }A } \)
\(=\frac { 2sinA }{ 1-co{ s }^{ 2 }A } =\frac { 2sinA }{ si{ n }^{ 2 }A } \) = 2cosecA

Now,cos\(\theta \) + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) 
Squaring both sides,
(cos\(\theta \) + sin\(\theta \) )² =(\(\sqrt { 2 } \) cos\(\theta \) )²
cos² \(\theta \) + sin² \(\theta \) + 2sin\(\theta \) cos\(\theta \)  = 2cos²\(\theta \) 
2cos²\(\theta \) - cos²\(\theta \) - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
cos²\(\theta \)  - sin²\(\theta \)  = 2sin\(\theta \) cos\(\theta \) 
(cos\(\theta \) + sin\(\theta \) ) (cos\(\theta \) + sin\(\theta \) ) = 2sin\(\theta \) cos\(\theta \) 
cos\(\theta \) - sin\(\theta \)  = \(\frac { 2sin\theta cos\theta }{ cos\theta +sin\theta } \) =\(\frac { 2sin\theta cos\theta }{ \sqrt { 2 } cos\theta } \) [since cos\(\theta \)  + sin\(\theta \) =\(\sqrt { 2 } \) cos\(\theta \) ]
=\(\sqrt { 2 } \) cos\(\theta \) 
Therefore cos\(\theta \) - sin\(\theta \)  =\(\sqrt { 2 } \) cos\(\theta \)

sin² Acos² B + cos² Asin² B + cos² A + cos² B+  sin² Asin² B
= sin² Acos² B + sin² Asin² B + cos² A + cos² B + sin²Asin² B
= sin² A(cos² B + sin² B) + cos² A(sin² B + cos²B)
= sin² A(1) + cos² A(1) (since sin² B + cos² B = 1) 
= sin² A + cos² A = 1